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  • Class 11th Computer Chapter 4: Applications and uses of Computer

    11th Class Chapter 4: Applications and uses of Computer Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Computer Chapter 4 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Computer Chapter 4 User Interaction are available to download or view. Many students practice 2024 Computer Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q1. Computer Simulation and Weather Forecast?

    (a) Uses of Computer Simulation?

    Ans.Computer simulation creates artificial models of systems, helping with training, environmental familiarization, and educational understanding. For instance, it’s used in pilot training, allowing students to conduct complex experiments.

    Q2. How can computers be useful in weather forecasting?

    Ans.Computerized weather forecasting systems gather data from various sources, including weather stations, airports, and satellites. They process this data to generate weather forecasts. This technology aids in predicting complex weather patterns and is used by organizations like the SPARCO weather forecasting department.

    (a) List three uses of computer in E-commerce.

    1. Electronic Shopping (E-Shopping): Customers can buy goods and services online from businesses, making use of computers from home, work, or cyber cafes.
    2. Video Conferencing: Businesses conduct virtual meetings between people at different locations, saving time and travel costs.
    3. Electronic Banking: Customers can perform online banking transactions from home, work, or while on the go, eliminating the need to visit physical bank locations.

    Q4. How can computers be useful in business?

    Ans. Computers are essential in the global business landscape. They expedite various business processes, reduce administrative paperwork, and enable online sales and customer communication

    Q5. How can computers be useful in education?

    Ans. Computers revolutionize education by allowing students to solve problems efficiently and access vast information online. They are used in colleges and universities through methods like Computer-Aided Learning (CAL), Computer-Based Training (CBT), and Online Education, enhancing teaching and learning experiences.

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    For the 11th Class Computer Chapter 4 Applications and uses of Computer, this set of notes follows the new syllabus, as it is for all Punjab boards. Other boards offer notes that different from this set. Faisalabad Board, Gujranwala Board, Rawalpindi Board, Sargodha Board, DG Khan Board, Lahore Board, Multan Board, Sahiwal Board, AJK Board are some of the boards in Punjab.

    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.


  • Class 11th Computer Chapter 3: Data communications

    11th Class Chapter 3: Data communications Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Computer Chapter 3 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Computer Chapter 3 User Interaction are available to download or view. Many students practice 2024 Computer Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q1. What is data communication?

    Ans.Data communication is the process of electronically transferring data from one place to another, facilitated by transmission media and communication devices.

    Q2. What are the basic elements of data communication?

    Ans. The fundamental components of data communication include sending and receiving devices, communication devices, and the transmission medium.

    Q3. What is communication media in data communication?

    Ans. Communication media, also known as communication channels, are the paths through which data is transmitted. Examples include telephone lines, twisted pair wires, coaxial cables, fiber optic cables, microwave, and satellites.

    Q4. What are the two main types of communication media?

    Ans. Communication media are categorized into two main types: physical transmission media (bounded media) and wireless transmission media (unbounded media).

    Q5. Can you explain physical transmission media with examples?

    Ans. Physical transmission media involve directly connecting communication devices with physical media. Examples include twisted pair wires (used in LAN), coaxial cables (used by cable TV and telephone companies), and fiber optics (used by telephone companies, ISPs, and cable TV operators).

    Q6. Describe wireless transmission media and provide examples?

    Ans.Wireless transmission media allow communication devices to exchange data through the air or space, using radio signals, microwave signals, and infrared signals. Examples include microwave transmission (utilizing high-frequency radio waves), communication satellites (space stations sending and receiving data), and mobile communication (radio-based networks for mobile computers).

    Q7. What is modulation in data communication?

    Ans.Modulation is the process of converting digital signals into analog form to facilitate data transmission. It combines data or message signals with carrier waves or signals, allowing information to be transmitted through space as electromagnetic waves.

    Q8. What are three basic methods of modulation, and how do they work?

    Ans. Three fundamental modulation methods are:

    1. Amplitude Modulation (AM): Modifies the carrier wave’s amplitude to represent binary digits 1 (presence of the carrier wave) and 0 (absence of the carrier wave). Commonly used in AM radio broadcasts.
    2. Frequency Modulation (FM): Alters the carrier wave’s frequency to represent binary digits 0 (original carrier frequency) and 1 (higher frequency). Commonly used in radio transmissions.
    3. Phase Modulation (PM): Varies the phase of the carrier wave to represent binary digits 1 and 0, similar to frequency modulation.

    Q9. What are the basic elements of a data communication system?

    Ans. The fundamental elements of a data communication system include:

    1. Sender/Transmitter (e.g., computer or telephone)
    2. Message (data or information to be transmitted)
    3. Medium or Communication Channel (path for data transmission, e.g., telephone line or fiber optics)
    4. Receiver (e.g., computer, telephone, or printer)
    5. Encoder and Decoder (used to convert signals between different forms)

    Q10. How do asynchronous and synchronous transmissions differ?

    Ans. Asynchronous Transmission:

    • Data is sent character by character.
    • Data is not stored before transmitting.
    • May have gaps between transmitted characters.
    • Uses start and stop bits to control information transmission. Synchronous Transmission:
    • Data is transmitted block by block or word by word.
    • Data is stored before transmission.
    • No gaps between transmitted characters.
    • Relies on clock signals for information transmission control.

    Q11. What’s the difference between Narrow Band and Broad Band in data communication?

    Ans.Narrow Band:

    • Low-speed transmission, typically 40 to 300 bps or baud.
    • Used for low-speed teletypewriter communication and similar devices. Broad Band:
    • High-speed transmission, capable of rates around 56,000 bps or baud.
    • Utilized when significant amounts of data need to be transmitted at high speeds.

    Q12. What is meant by Simplex and Half-Duplex modes of communication?

    Ans. Simplex Mode:

    • Data flows in only one direction; unidirectional.
    • Devices can either send or receive data but not both simultaneously.
    • An example is traditional television broadcasting. Half-Duplex Mode:
    • Data can flow in both directions but not simultaneously.
    • Devices can send or receive data but not at the same time.
    • Internet surfing, where users send requests and receive web pages one at a time, is an example of half-duplex communication.

    Q12. Why is Fiber Optic cable used for communication?

    Ans.Fiber Optic cable is chosen for communication because it offers high-speed data transmission (usually 1Gbps to 2Gbps) and is not affected by electromagnetic interference. Its optical properties, like refraction, enable the transmission of data as light waves. Many telephone and cable TV companies employ fiber optic cables for communication, making it a preferred choice for modern communication systems.

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    For the 11th Class Computer Chapter 3 Data communications, this set of notes follows the new syllabus, as it is for all Punjab boards. Other boards offer notes that different from this set. Faisalabad Board, Gujranwala Board, Rawalpindi Board, Sargodha Board, DG Khan Board, Lahore Board, Multan Board, Sahiwal Board, AJK Board are some of the boards in Punjab.

    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.

  • Class 11th Computer Chapter 2: Information Network

    11th Class Chapter 2: Information Network Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Computer Chapter 2 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Computer Chapter 2 User Interaction are available to download or view. Many students practice 2024 Computer Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q1. Describe Computer Network?

    Ans. A computer network is a collection of computers and devices linked together through communication channels, allowing them to share information and resources like printers and hard disks. This interconnection can be achieved through various technologies, including cables, telephone lines, radio waves, and satellites. The internet is one of the most popular examples of a computer network.

    Q2. Distinguish between Dedicated Server Network and Peer-to-Peer Network?

    Ans. Dedicated Server Network:

    • In a dedicated server network, computers are classified as either servers or clients.
    • One or more computers act as dedicated servers that control the network.
    • Data and program files are primarily stored on the dedicated server, making data management efficient.
    • Often used in large organizations for enhanced security.

    Peer-to-Peer Network:

    • In a peer-to-peer network, all computers are peers, acting as both servers and clients.
    • There is no dedicated server or central computer controlling the network.
    • Each computer stores its data and program files, making data management less efficient.
    • Suitable for small offices and offers lower security.

    Q3. Define Workgroup Computing?

    Ans. Workgroup computing refers to a group of individuals working together on a specific task and sharing information through a computer network. It is a form of collaborative computing where members of the group can access and exchange information over the network.

    Q4. Two Disadvantages of Workgroup Computing?

    1. Ans. Setting up individual accounts and passwords on each machine can be time-consuming. Passwords may become out of sync if changed on one computer and not on others.
    2. Workgroup computing is not easily scalable. As the number of computers increases beyond a certain point (e.g., more than 10), setting up accounts for new users becomes increasingly time-consuming.

    Q5. Define Network Topology?

    Ans. Network topology refers to the way computers and nodes are connected within a computer network. It defines the arrangement and structure of the network, specifying how devices communicate and interact with each other.

    Q6. Explanation of Types of Network Topologies?

    Ans. There are several common network topologies, including:

    1. Bus Topology: All nodes connect to a central communication medium (bus).
    2. Star Topology: All nodes connect directly to a central device, such as a hub or switch.
    3. Ring Topology: Nodes are connected in a circular chain.
    4. Mesh Topology: Each node is connected to every other node, offering redundancy and robustness.
    5. Tree Topology: Combines characteristics of both bus and star topologies, forming a hierarchical structure.

    Q7. Comparison of BUS and STAR Topologies?

    Ans. Bus Topology:

    • Nodes connected to a common communication medium (bus).
    • Data and destination addresses sent through the bus.
    • Data moves from one computer to another until it reaches the destination.

    Star Topology:

    • Nodes directly connected to a central device (hub/switch).
    • Sending computer sends data to the hub, which then forwards it to the receiving computer.

    Q8. Three Benefits of Networks?

    1. Ans. Communication: Networks enable cost-effective communication through methods like email, chat, telephone, and video conferencing.
    2. Data Sharing: Users can share data and information easily among authorized network users.
    3. Resource Sharing: Resources like printers, scanners, hard disks, and modems can be shared among different network users.

    Q9. Define Computer Bus and Its Functions?

    • Ans. Computer Bus: A computer bus is a set of electrical paths that enable the CPU to send and receive data and instructions to and from various computer components. It has a specific bus width, indicating the amount of data it can carry at once.
    • Functions of Computer Bus: A computer bus connects all devices within a computer system. Its speed, known as bus clock speed, affects overall computer performance. Bus clock speed is measured in megahertz (MHz).

    Q11. Explanation of OSI, Protocols, Network Layer, and Circuit Switched Network?

    • Ans. OSI (Open System Interconnection): OSI is a complete model for network communication that covers all aspects of network communications. It consists of seven layers, each with specific functions.
    • Protocols: Protocols are sets of rules for exchanging data between computers on a network. They define what, how, and when data is communicated. Examples include TCP/IP, SLIP, PPP, HTTP, and FTP.
    • Network Layer: The network layer is responsible for establishing, maintaining, and terminating network connections. It manages data delivery from source to destination, determining logical paths between sender and receiver. Common protocols operating at this level are IP, IPX, and X.25.
    • Circuit Switched Network: In a circuit-switched network, a dedicated connection is established between two nodes before they can communicate. A temporary circuit is set up, and it remains connected for the duration of the communication session, mimicking a physical circuit. Telephone networks are an example of circuit-switched networks.

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    For the 11th Class Computer Chapter 2 Information Network, this set of notes follows the new syllabus, as it is for all Punjab boards. Other boards offer notes that different from this set. Faisalabad Board, Gujranwala Board, Rawalpindi Board, Sargodha Board, DG Khan Board, Lahore Board, Multan Board, Sahiwal Board, AJK Board are some of the boards in Punjab.

    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.


  • Class 11th Computer Chapter 1: Fundamantals of Internet

    11th Class Chapter 1: Fundamantals of Internet Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Computer Chapter 1 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Computer Chapter 1 User Interaction are available to download or view. Many students practice 2024 Computer Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q1. What is Information Technology (IT)?

    Ans. Information Technology (IT) is the use of computing and high-speed communication to share information globally, connecting computers for data exchange.

    Q2. Differentiate between System Software and Application Software?

    • Ans. System Software: Controls computer operations and enables other applications. Examples include operating systems and utility programs.
    • Application Software: Solves specific user problems and is divided into custom-built and packaged software.

    Q3. Differentiate between Dot Matrix and Daisy Wheel Printers?

    • Ans. Dot Matrix Printer: Prints characters one at a time using pins in a matrix.
    • Daisy Wheel Printer: Resembles a typewriter with a circular print wheel.

    Q4. Define Memory Cell, Memory Word, and Byte?

    Ans. Memory Cell: An electronic switch that stores 0 or 1.

    • Memory Word: A group of bits representing data.
    • Byte: A group of 8 bits.

    Q5. Write a short note on a plotter?

    Ans. A plotter is an output device used for large-format printing of architectural drawings, maps, and charts. It works by moving a pen on paper to draw lines.

    Q6. Define the working of a scanner?

    Ans. A scanner is a light-sensing input device that scans images with light, converting them into digital form for storage in a computer.

    Q7. Brief Note on Non-impact Printers?

    Ans. Non-impact printers, like Laser, Inkjet, and Thermal printers, produce characters and graphics on paper without physically striking it. They use methods such as lasers, heat, and pressure to create prints. These printers are faster, quieter, and offer better print quality compared to impact printers.

    Q8. Definition of Static and Dynamic Memory Devices?

    • Ans. Static Memory Devices: These are semiconductor memory devices that don’t need periodic refresh. They are faster than dynamic memory devices. Example: SRAM.
    • Dynamic Memory Devices: These semiconductor memory devices require periodic refreshing, making them slower. Example: DRAM.

    Q9. Short Note on Computer Software?

    Ans. Computer software is a set of instructions given to a computer to solve specific problems. It tells the computer what to do and how to do it, enabling various operations. Software is developed by writing instructions in a programming language.

    Q10. Description of RAM and Its Types?

    Ans. RAM (Random Access Memory) allows direct access to individual bytes and temporarily stores data and instructions. It’s volatile, meaning its contents are lost when the power is turned off. RAM has two types: DRAM (Dynamic Random Access Memory) and SRAM (Static Random Access Memory).

    Q11. Explanation of Access Time in RAM?

    Ans. Access time in RAM is the time taken by a processor to read or write data, measured in nanoseconds. Faster access time leads to quicker processing. For fast RAM chips, the access time is typically around 70 nanoseconds.

    Q12. Brief on System Development in Information System Development?

    Ans. System Development involves writing program code in a programming language. It’s a crucial phase where programmers and software engineers use program specifications, algorithms, and flowcharts to create software.

    Q13. Definition of Cycle Time?

    Ans. Cycle time is the duration needed to perform a read or write operation on a memory location and return to its original state for the next operation. It’s measured in nanoseconds and is typically longer than access time.

    Q14. Difference Between Random Access Memory and Sequential Access Memory?

    Ans. Random Access Memory (RAM): Allows the CPU to perform random read and write operations. It’s read/write memory, very fast, with the same access time for any memory location.

    Sequential Access Memory: Enables the CPU to perform read and write operations sequentially, reading and storing information one after the other. It’s slower, with varying access times for different locations.

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    For the 11th Class Computer Chapter 1 Fundamantals of Internet

    , this set of notes follows the new syllabus, as it is for all Punjab boards. Other boards offer notes that different from this set. Faisalabad Board, Gujranwala Board, Rawalpindi Board, Sargodha Board, DG Khan Board, Lahore Board, Multan Board, Sahiwal Board, AJK Board are some of the boards in Punjab.

    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.

  • Class 11th Chemistry Chapter 8: Introduction TO Chemical kinetics

    11th Class Chapter 8: Introduction TO Chemical kinetics Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Chemistry Chapter 8 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Chemistry Chapter 8 User Interaction are available to download or view. Many students practice 2024 Chemistry Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q.1: Define activation energy and activated complex?

    Ans: Activation energy is the maximum amount of energy more than the average energy, which is just sufficient convert reactants into products.
    Activation complex is the collection of molecules of reactions as the time of collision having greater energy than the reactions and products. At this stage old bonds become
    weak and the possibilities for the new bonds are then.

    Q.2: What is temperature coefficient of a reaction?

    Ans: It has been observed that by increasing the temperature of a reaction by 10 K. the rate constant doubles. This is called temperature coefficient of a reaction. Bagsp

    Q.3: What is the effect of temperature on tire activation energy of a reaction?

    Ans: Activation energy of the chemical reactions is a constant parameter and does not depend upon the temperature. Any how, by increasing the temperature greater number of molecules get a chance to cross the activation energy barrier.

    Q.4: The rate of a chemical reaction is an ever changing parameter under the given conditions?

    Ans: When the reaction progresses, the reaction is very fast at the beginning, slow somewhere in the middle and very very slow at the end. The reason is that rate depends upon the concentration according to the law of mass action. The Concentrations decreases every moment, so rate decreases every moment.

    Q.5: The reaction rate decreases every moment but the rate constant ‘K’ of the reaction is a constant quantity, under the given conditions? Justify it?

    Ans: Rate constant is the rate of chemical reaction when the concentration of reactants are unity.
    A+B C+D
    The concentration of A and B decreases with the passage of time, so the rate decreases, but the rate constant “K” remains the same for the reactions throughout, its progress under the given conditions.

    Q.6: When the reaction become zero order?

    Ans: Those chemical reactions whose rates are independent of the concentrations of reactants and are mostly influenced by a suitable catalyst, or light are zero order reaction. Thus units for rate constant for such reactions are equal to the units of rates of reactions.

    Q.7: The radioactive decay is always a first order reaction? How?

    Ans: Radioactive substance have a single species at a moment whose nucleus is being broken up without the help of any external agency. So only one reactant is present and it allows the first order mechanism.

    Q.8: What type of information is obtained for the order of reaction from half life period of that reactions?

    Ans: Half life period depends upon the initial concentration of reactants. Actually, half life period is inversely proportional to the initial concentration of reactants raised to the power one less than the order of reaction. So from the initial concentration and half life period order of reactions can be calculated.

    Q.9: The reaction happened due to collisions among the molecules, but all the collisions are not fruitful?

    Ans: In order to do the chemical reaction, a minimum amount of energy, more than the average energy is required. Hence only those collisions will prove to be fruitful whose energies are more than or equal to the energy of activation. Moreover, proper orientation is also required. Due to these facts the reactions do not happen in a single moment, but they takes some times.

    Q.10: How the energy activation can be calculated from the Arrhenius plots?

    Ans: When a graph is plotted between inverse of temperature on x-axis and log k on the y- axis, then a straight line with the negative slope is obtained. The slope of this straight line gives us the energy of activation.

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    For the 11th Class Chemistry Chapter 8 Introduction to Programming, this set of notes follows the new syllabus, as it is for all Punjab boards. Other boards offer notes that different from this set. Faisalabad Board, Gujranwala Board, Rawalpindi Board, Sargodha Board, DG Khan Board, Lahore Board, Multan Board, Sahiwal Board, AJK Board are some of the boards in Punjab.

    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.

  • Class 11th Chemistry Chapter 7: Solution & Electrolytes

    11th Class Chapter 7: Solution & Electrolytes Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Chemistry Chapter 7 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Chemistry Chapter 7 User Interaction are available to download or view. Many students practice 2024 Chemistry Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q.1: Why the boiling point of a solution of a non-volatile solute in a volatile solvent is always greater than the boiling point of a pure solvent?

    Ans: The elevation of the boiling point of the solution is due to the lowering of vapour pressure of the solution as compared to the pure solvent. In the case of solution the vapour pressure becomes less. In order to equalize the vapour pressure of the solution to the external pressure, more temperature us required than that in the pure suite of the solvent. Which causes elevation of boiling point.

    Q.2: Why a non-volatile solute in a volatile solvent lowers vapour pressure of solution?

    Ans: The particles of the solute are distributed throughout the bulk of the solution and some of the particles of the solute are also present on the surface of the solution. The number of molecules of the solvents per unit area on the surface of the solution become less. Hence the evaporating tendency of the solvent decreases and vapour pressure of solution becomes less.

    Q.3: Why the freezing points are depressed due to presence of solute?

    Ans: The lowering of vapour pressure compels the solutions to freeze at those temperature, which are below the freezing point of pure solvent. The reason is that the vapour pressure temperature curve meets the solid phase of pure solvent at lower temperature than the pure solvent.

    Q.4: Boiling poi of solvents increase due to the pressure of solute. Why?

    Ans: The surface of the solution has molecules of solute as well. They do not allow the solvent to leave the surface as rapidly as in pure solvent. The boil the solutions we have to increase the temperature of solutions in comparison to pure solvents.

    Q.5: Define Colligative properties? Name some important Colligative properties?

    Ans: Those properties of solutions, which depend on the number of the properties. These properties are:
    ➤ Lowering of vapour pressure
    ➤ Elevation of boiling point
    ➤ Depression of freezing point
    ➤ Osmotic pressure

    Q.6: What is the effect of temperature on the conjugate solution of water and phenol?

    Ans: Water and phenol are partially miscible. They make two distinct solutions and two separate layers at room temperature. By increasing the temperature, the composition of the two conjugate solution change and at 65.9°C a homogeneous mixture of two components is formed.

    Q.7: What is conjugate solution?

    Ans: When two partially miscible liquids are mixed, two layers are produced, They are the solutions of one component into the other. These two solutions have distinct line of demarcation and are called conjugate solutions.

    Q.8: What are the hydrates? How are they formed?

    Ans: The crystalline substances that contain chemically combined water in definite proportional
    is called a hydrate. They are produced when aqueous solution of soluble salt is evaporated.

    Q.9: How the forces of attraction between the ions and the solvent molecules depend upon the charge densities of the ions?

    Ans: Smaller the size of the ion or greater the amount of the positive charge on the ion, greater the charge density. Such ions have greater forces of attractions. For the molecules of the solvent. They are solved more effectively. Negatively charged ions are bigger in size, so that are not solvated in a better way.

    Q.10: How the ions are stabilized when a strong electrolyte like NaCl and KCI are dissolved in H₂O?

    Ans: Strong electrolyte area dissolved to the maximum extent and positive and negative ions are produced. These ions are surrounded by the water molecules as follows. Their charged are satisfied by the solvent molecules. In this way, they get the stabilities. INT

    Q.11: Heat lution of a substance Is measured at infinite dilution? Why?

    Ans: Actually the maximum amount of heat is evolved or absorbed, when each particle of one mole of solute is completely solved by the solvent molecules. These solvated particles of the solute should be far away from each other in the solution. This is only possible when this solution is very dilute.

    Q.12: When the heat of solution in negative, then increase in temperature decreases the solubility and vice versa? Why?

    Ans: When the heat of solution is negative it means that the vessel is heated up during the solution formation. So when heat is supplied from outside, then the system will go to that side where greater amount of heat can be stored and that is the side of low solubility.

    Q.13: Why the NaCl and KNO, are used to lower the melting point of ice?

    Ans: NaCl and KNO, are electrolytes and are sufficiently soluble in water. They double the number of particles after dissociation in water. In this way, they can manage to decrease the freezing point of water to a greater extent as compared to a non-electrolyte.

    Q.14: The concentration in terms of molarity is independent of temperature, but morality depends upon temperature? Why?

    Ans: In molal solutions the mass of the solvent and that of the solute are also fixed. The masses of the substances are not temperature dependent. In molar solutions are have the volume of solutions. Volume of a liquid is a temperature dependent. So the molality is not influenced by temperature but molality does change.

    Q.15: The total volume of the solution by mixing 100 cm³ of water with 100 cm³ of alcohol may not be equal to 200 cm³? Justify it?
    Ans: In one molal solution of urea, 60 grams of urea is dissolved in 1000 g of water, which is approximately 1000 cm³ of water. In one molar solution of urea, 60 g of urea is added in water to make total volume of solutions as 1000 cm³. So the volume of water in molar solution is concentrated and molal solution is dilute.

    Q.16: Why glucose is not soluble in CCl4, but dissolve in water?

    Ans: The molecule of glucose has strong hydrogen bonding due to the presence of give -OH groups, CCI, being non polar molecule solvent, cannot break the hydrogen bond ing among glucose molecules water is polar solvent, so it breaks the hydrogen bonding of glucose and dissolves it.

    Q.17: What do you mean by discontinuous solubility curves?

    Ans: The graph between solubilities and temperature may show a sudden change in the direction. Such a graph is called discontinuous solubility curve. Such curve is the combinations of two or more than two curves due to the changing behaviours of that compound.

    Q.18: How does fractional crystallization help in removing the impurities from a solid substance?

    Ans: The impure substance is dissolved in hot solvent. The solute to be purified should be less soluble than impurities. When cooling is done, this solute settle down in the form of crystals and impurities are left behind in the solution. Filtration is done to get the crystals of the required substances.

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    For the 11th Class Chemistry Chapter 7 Introduction to Programming, this set of notes follows the new syllabus, as it is for all Punjab boards. Other boards offer notes that different from this set. Faisalabad Board, Gujranwala Board, Rawalpindi Board, Sargodha Board, DG Khan Board, Lahore Board, Multan Board, Sahiwal Board, AJK Board are some of the boards in Punjab.

    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.

  • Class 11th Chemistry Chapter 6: Chemical Equilibrium

    11th Class Chapter 6: Chemical Equilibrium Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Chemistry Chapter 6 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Chemistry Chapter 6 User Interaction are available to download or view. Many students practice 2024 Chemistry Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q.1: Justify that chemical equilibrium is dynamic in nature?

    Ans: In reversible chemical reactions the molecules of reactions collide and convert into products. At the same time the molecules of the products are converted into reactants. When two opposing forces maintain the equal, rates then equilibrium is there and that is dynamic equilibrium in nature.

    Q.2: When does four types of chemical equilibrium constants for a reaction become equal?

    Ans: When we have such a reversible reaction, in which the total number of moles of reactants and products are equal then Kp = Kc = Kn. In order words An = 0, where An is the difference of number of moles of reactants and products.

    Q.3: The change of temperature disturbs the equilibrium position and the equilibrium constant of reaction? Justify?

    Ans: All the reversible reactions are disturbed by changing their equilibrium position and equilibrium constant by disturbing the temperature. Actually change of temperature changes the energy contents of reactants and products.

    Q.4: Why is solubility of cose in water is increased by increasing the temperature?

    Ans: The solubility of glucose in water is an endothermic process. Increase of temperature pushed the system to that side where heat is absorbed, and that is the side where solubility increases.

    Q.5: What the nature of solution having PH = 7.5?

    Ans: The solution with PH 7 is neutral and having values more than 7 have greater concentration of OH and less concentration of H*, so the solution is basic.

    Q.6: What is buffer capacity?

    Ans: Buffer capacity is a ability of a buffer to resist the change of PH, when a few drops of an acid and base is added from outside. The best buffer is obtained. (i) when PH of the buffer is equal to Pka of the acid and the concentration of the salt and acid are equal. (ii) The OH of the buffer is equal to Pkb or the base.

    Q.7: Why do we need buffer solution?

    Ans: Buffer solutions are used in many industrial processes as electroplating manufacture of leather, preparation of dyes and manufacture of photographic materials. Buffer solutions are used by analytical chemists and to calibrates PH meter. They are also in culture media.

    Q.8: How does a catalyst affect a reversible reaction?

    Ans: A catalyst affects the rates of both steps equally, so the equilibrium position remains the same. It means that equilibrium constant should be the same. Actually a catalyst decreases the energy of activation of the chemical reaction by giving a new path to the reaction. In this way, a greater % age of reactant molecules is able to cross the energy barrier and the rate of reaction increases.

    Q.9: What is meant by stage of chemical equilibrium?
    Ans: In reversible chemical reactions two opposing reactions occur. A stage reaches for the reaction when the rates of two opposing reactions are equal. This stage is called stage of chemical reaction.

    Q.10: Why the rates of chemical reaction is depend upon the concentrations of reactants?

    Ans: This is according to the law of mass action. The rates of chemical reactions are directly proportional to the product to the active masses of reactants. Greater the number of molecules, greater the collisions and greater the rate.

    Q.11: Why is the equilibrium blished? Only the reaction vessel is closed for a gaseous reaction?

    Ans: If some of the products or the reactants in the open vessel leave the vessel during the progress of the reaction then the stage of equal rates in distributes and equilibrium can not be established. In such situation the reaction is pushed to the toward direct ion or backward direction.

    Q.12: The change of volume disturbs the equilibrium position for some of the gaseous phase reactions but not the equilibrium constant. Why?

    Ans: Those gaseous phase reversible reactions, which happens with changing number of moles are affected by the change of volume at equilibrium stage. Their equilibrium position is disturbed, but equilibrium is not changed.

    Q.13: What is the effect of rise in temperature on the stability of Kl in water?
    Ans: The heat of solution of KI in water in positive. It means it is an endothermic process. When the temperature is increased, it will absorb more heat and more dissolution will take place.

    Q.14: What is buffer solution?

    Ans: A solution which resists the change of PH, when a small amount of an acid or a base is added in that. Buffer are prepared by mixing two components. Weak acid + salt with a strong base. Weak base + salt with a strong acid.

    Q.15: Define the effect of common ion solubility? Give examples?

    Ans: According to the Le Chatelier’s principle, if a common ion is added in a solution, then the solubility of the electrolyte decreases and the solute is completed to settle down. If Cl are added in saturated solution of NaCl, then NaCl settles down, because its solubility decreases.

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    For the 11th Class Chemistry Chapter 6 Introduction to Programming, this set of notes follows the new syllabus, as it is for all Punjab boards. Other boards offer notes that different from this set. Faisalabad Board, Gujranwala Board, Rawalpindi Board, Sargodha Board, DG Khan Board, Lahore Board, Multan Board, Sahiwal Board, AJK Board are some of the boards in Punjab.

    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.

  • Class 11th Chemistry Chapter 5: Energetics of Chemical Reaction

    11th Class Chapter 5: Energetics of Chemical Reaction Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Chemistry Chapter 5 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Chemistry Chapter 5 User Interaction are available to download or view. Many students practice 2024 Chemistry Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q.1: How the lattice energy of the ionic compound can be measured by Born Haber cycle?

    Ans: We first of all determine heat of formation of compound in laboratory. The heat of sublimation, heat of ionization, amount of electron affinity, bond dissociations energy of the substances are equalized with heat of formation.

    Q.2: State the first law of Thermodynamic and give its mathematical form?

    Ans: According to the first law of thermodynamic. energy can neither be created nor destroyed. If a gas is enclosed in a piston cylinder arrangement and it is heated by giving some calories of energy, then the piston can be pushed to the backward direction. So some of the heat is converted into internal energy and others used up to do the work.

    Q.3: How do we determine the AH in the laboratory for food, fuel etc.?

    Ans: For this purpose, bomb calorimeter is used. The substance under test is ignited electrically in the atmosphere of oxygen. The amount o at produced is calculated from the mass of the substance, specific heat of the calorimeter and change of temperature.

    Q.4: The total energy of a system is the sum of translational, rotational and vibrational motions? Justify it?

    Ans: The substances are consisted of atoms, ions and molecules. If a system is in liquid or gaseous state then its constituent particles have translational motion along three axis x, y and z space. They need energy for this. For rotational motion along three axis they also need energy. Various types of vibrational motions of the bonds make the bonds to compress and stretch. They also demand energy. So the total energy of a system is a sum of all these three types of energies.

    Q.5: How the temperature of the system changes during exothermic and endothermic reactions?

    Ans: In an exothermic reaction, heat is evolved with increases the temperature of the system. In an endothermic reaction, heat is absorbed so the temperature of the system falls down. These statement are true when the system is isolated.

    Q.6: What is a spontaneous process?

    Ans: That process which takes on its own is called spontaneous process. No external assistance is required. It moves from non-equilibrium state towards the equilibrium state. It is unidirectional and irreversible.

    Q.7: What is state and state function?

    Ans: The condition of a system is called state of a system. The initial state is a description before a change happens. Final state is a description after a change takes place. State function is the macroscopic property of a system, which has some definite values for initial and final states and is independent of the path which is adopted to bring out the change.

    Q.8: Define the following terms:
    ➤ Enthalpy
    ➤ Endothermic reaction.
    Ans: Enthalpy is the total heat content of the system. It cannot be measured. We can only measure the enthalpy changes of a system. Those chemical reactions whose enthalpy is positive and they absorb the heat during the process of the reaction are called endothermic reaction.

    Q.9: What is the physical signifwance of equation AH = qp?

    Ans: AH is the enthalpy change and qp is amount of heat supplied at constant pressure. This equation shows that the expansion of the gas takes place and work is done by the system.

    Q.10: How will you differentiate between AE and AH? Is it true that AH and AE have the same value for the reactions taking place in the solution state?

    Ans: AH is the heat change at constant pressure, while AE is the heat change at constant volume. In the solution state the pressure is constant, So, there is no difference in AE, and AH in such solution. There is no change in volume when the reaction happens in liquid or solid state.
    Hence
    PΔV = 0
    ΔH = ΔE + PΔV
    ΔΗ = ΔΕ + P Χ Ο
    ΔH = ΔE

    Q.11: Specific heat of a substance depends upon the nature of substance? Why?

    Ans: Specific heat is the amount of heat required to raise the temperature of 1 gram of substance by 1 kelvin. The specific heat of H2O is 42 Jg¹k, while for ethanol it is 2.4 Jg¹k¹. Actually it is the capability depends upon the structure, bond angles and vibrational frequencies of the bonds etc.

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    For the 11th Class Chemistry Chapter 5 Introduction to Programming, this set of notes follows the new syllabus, as it is for all Punjab boards. Other boards offer notes that different from this set. Faisalabad Board, Gujranwala Board, Rawalpindi Board, Sargodha Board, DG Khan Board, Lahore Board, Multan Board, Sahiwal Board, AJK Board are some of the boards in Punjab.

    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.

  • Class 11th Chemistry Chapter 4: Chemical Bonding

    11th Class Chapter 4: Chemical Bonding Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Chemistry Chapter 4 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Chemistry Chapter 4 User Interaction are available to download or view. Many students practice 2024 Chemistry Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q.1: What is bond? Define the mean types of bonds?
    Ans: DEFINITION:

    The attraction forces which bind the two or more atoms (like or unlike) molecule of a compound is called chemical bond.
    TYPES OF BOND:

    There are three types of chemical bond defined the follows.
    Ionic Bond:

    It is formed by transference of electron between two atoms also known as electrovalent bond. Such a bond which is formed by the electrostatic attraction between positive and negative ion is called ionic or Electrovalent bond.
    Covalent Bond:

    It is formed between two atoms by the mutual sharing of electrons is known as covalent bond.
    Coordinate Covalent Bond:

    Formed by one sharing of electrons (in which one atom provides shared pair of electrons)

    Q.2: whta is single convalent bond?

    Ans: Which is formed by the mutual sharing of one electron pair between two atoms e.g. HCIH-CL-HCL

    Q.3: What is double covalent bond?

    Ans: Which is formed between two atoms by the mutual sharing of two electrons pairs and denoted by double bond (=) e.g. 0:00 0

    Q.4: What is triple covalent bond?

    Ans: Covalent bond formed by mutual sharing of three electron pairs also known as triple covalent bond denoted by triple bond. e.g.

    Q.5: Define Dipole Moment?

    Ans: DEFINITION:
    Dipole moment of a molecule is defined as the vector equal in magnitude to the product of the electric charge “q” and the distance “r” having the direction of the line joining positive and negative centres.
    Mathematical representation
    Dipole moment = charge x distance
    i.e.µ = qxr

    Q.6: What is Bond Energy?

    Ans: It is the amount of energy released or the amount of energy required during the formation or breaking of the bond between two atoms respectively. e.g. CICI has B.E = 2.44 kj/mole.

    Q.7: What is hybridization?

    Ans: The term hybridization means blending or mixing of pure atomic orbitals of different energies and shape to give equal number of hybrid orbitals having the same energy and shape. This newly produced orbitals are known as hybridized or hybrid orbitals.

    Q.8: What is hydrogen bonding?

    Ans: The attractive force which combines the electropositive hydrogen atom of one molecule with the electronegative atom of the other molecule is known as “hydrogen bond”. To distinguish hydrogen bond it is represented by dotted line (……..). This is known “protonic bridge”. e.g. Hydrogen bonding in HF Practicaentre
    H&+ – F&-
    Hydrogen bond in HF
    H&+ – F8-

    Q.9: What is the shape of water molecule as per Lewis structure?

    Ans: The shape of water molecule as per lewis structure Hu central oxygen atom is surrounded by four active pairs.

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    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.

  • Class 11th Chemistry Chapter 3: The Atomic Structure

    11th Class Chapter 3: The Atomic Structure Short and Simple Question & Answer

    We are aware that you are searching for 11th Class  Chemistry Chapter 3 Notes on the internet. The notes are well-written, simple, and organized in an easy-to-understand manner and according to the new syllabus. At the bottom of these notes, you will find a download button to make your life easier. These notes for 11th Class Chemistry Chapter 3 User Interaction are available to download or view. Many students practice 2024 Chemistry Notes questions by FAIZ UL ISLAM and get good marks in the exam.

    Q.1: What are defects of Rutherford’s atomic model?

    Ans: Rutherford’s atomic model are:
    ➤ It is based on the laws of the motion and the gravitation. These laws charged bodies like electrons.
    ➤ The revolving electron should emit the energy continuously and by producing a spiral path, it should ultimately fall into the nucleus. But actually the atoms don’t do.
    The atom should go continuously spectra, but they give the line spectra.

    Q.2: How does Bohr introduce the Plank’s Quantum Theory in hic model?

    Ans: Bohr produced that electrons move around the nucleus in the fixed orbits with definite energies. Whenever, they change the orbits they eit or absorb the energy in terms of photons which was suggested by Planks.

    Q.3: Indicate the limiting line of Balmer series?

    Ans: When the electrons jumps from an infinite orbit to n = 2 of hydrogen atom then this line on the photographic plate is called the limiting line, it has energy and lies in the u.v region.

    Q.4: How did Rutherford’s model of an atom first af all proved the existence of nucleus of the atom

    Ans: Rutherford’s observed that most of the alpha particles passed straight through the gold foil without any deflection from the path. Few of them were deflected at some angles more than 900 and a few were deflected heavy and positively charged part at the center. This heavy part at the centre is called nucleus.

    Q.5: Why the anode rays depend upon the nature of the gas?

    Ans: Anode rays are those particles which are consisted of rest of the atom or molecule after the removal of one electron. The mass of every anode rays particle depends upon the nature of the gas, so the anode ray for all gaseous substances are different.

    Q.6: The e/m value of positive rays for different gases are different? But those for cathode rays the elm value are same? Justify?

    Ans: The nature of particles of positive rays in a discharge tube depend upon the nature of the gas because the nucleus of every gas has its own number of protons and neutrons.Greater the number of protons and neutrons in the nucleus of an atom or nuclei of the molecule, smaller the elm values. In this case of cathode rays which are always electrons. E/m values remain the same.

    Q.7: Define Hund’s Rule?

    Ans: This rule is concerned with the distribution of electrons in degenerate orbitals. These orbitals may be atomic or molecular. According to this rule if degenerate orbitals are available and more than one electrons are to be placed in them, then place them in separate orbitals with the same spin rather than putting them in the same orbital with opposite spin.

    Q.8: The magnetic quantum number gives us the orientation of orbital in space? Justify?

    Ans: In order to designed the direction of p-sub shell, we need an additional quantum number and that is called magnetic quantum number. It tells us the orientation of orbital in space p sub shell has the orbitals and they have three directions in spacer For each there s a separate value of magnetic quantum number.

    Q.9: What is the function of principle of quantum number?

    Ans: This quantum number () give us the shape of the orbital.
    When n = 0, the orbital is S
    When n = 1, the orbital is P
    When n = 2, the orbital is d
    When n = 3, the orbital is f

    Its value also determines the energy of electrons.

    Q.10: Heisenberg’s uncertainly principle has no relation with Bohr’s Atomic model? Justify it?

    Ans: Since the electron has wavy nature and paths are elliptical as well, so the simultaneous determination of position and momentum is not possible. But Bohr’s model does not accommodate the wavy nature of electron. He says that the paths are fixed orbits and their orbits are planer. It means that Bohr’s model is very simple as compared to Heisenberg’s uncertainly principle.

    Q.11: Explain Cathode rays are material particles having definite mass and velocities?

    Ans: Cathode rays are material particles having definite mass and velocity because they derive as small peddle wheel placed in their path.

    Q.12: Describe continuous spectrum?

    Ans: In this spectrum boundary line between the colours cannot be marked the colours and diffused into each other. It is obtained from sunlight or bulb light. Example of continuous spectrum is rainbow.

    Q.13: Define atomic spectrum?

    Ans: In this spectrum lines are separated by definite distance depending upon the nature of element.

    Q.14: How are x-rays produced?

    Ans: X-rays are produced when rapidly moving electron collide with heavy metal and in the discharge tube.

    Q.15: Give the characteristics of Electrons, protons and neutrons?

    Ans: CHARACTERISTICS OF ELECTRON:
    It is a negatively charged particle.
    Charge of electron is 1.6022 x 1019 coulombs.
    Mass of electron is 0.00548597 a.m.u. Or 1.1 x 10¹³¹ kg.
    Electron is represented by “e”. Electrons revolve around the nucleus of atom is different circular orbits.
    CHARACTERISTICS OF PROTON:
    Proton is a positively charged particle. Charge of proton is 1.6022 x 1019 coulombs.
    Mass of proton is 1.0072766 a.m.u. Or 1.6726 x 1027 kg.
    Proton is 1837 times heavier than an electron. Protons are present in the nucleus of atom.
    CHARACTERISTICS OF NEUTRON:
    It is a neutral particle because it has no charge.
    Mass of neutron is 1.0086654 a.m.u. Or 1.6749 x 10-27 kg.
    Neutron is 1842 times heavier than an electron.
    Neutrons are present in the nucleus of an atom.

    Q.16: How is wavelength of the electron related to momentum of electron?
    Ans: Wavelength of electron is related to the momentum of electron on
    y = h/mv
    where
    Am is the mass of particle
    V is the velocity of electron
    ……is the De borglie’s wavelength.

    Q.17: Cathode rays are deflected towards positively charged plate and cathode rays deflected towards negatively charged plate? Justify?

    Ans: Cathode rays consist of negatively charged particles and anode rays consist of positively charged particles. Because opposite charges attract each other therefore cathode rays are deflected towards positively charged plate and anode rays are deflected towards negatively charged plate.

    Q.18: Cathode rays produce shadow of an opaque object placed in their path? Explain?

    Ans: Because cathode is material in nature and consists of streams of particles which are not pass through opaque and form shadow.

    Q.19: Why the anode rays called canal rays?

    Ans: Because anode rays pass through canals (opening/cuts into cathode) and they are called canal rays.

    Q.20: Alpha rays have penetrating power than Beta rays?

    Ans: Alpha rays have high penetrating power than beta rays due to being relatively heavy particles.

    Q.21: Gamma rays have more penetrating power than alpha and beta rays? Justify?

    Ans: Gamma rays have high penetrating power on account of high velocity and non – material nature.

    Q.22: Beta have high velocity than alpha rays? Why?

    Ans: Because alpha particles are heavier than beta particles.

    Q.23: How positive rays are produced?

    Ans: Positive rays are produced when high speed cathode rays strike with the molecules of a gas enclosed in the discharged tube.

    Q.24: How the bending of the cathode rays in the electric and magnetic fields shows that they are negatively charged?

    Ans: Cathode rays are deflected towards the positive plate when electric field applied. It shows that cathode rays are negatively charged. When cathode rays are passed through the magnetic field, they bend perpendicularly to the joining line of two poles.
    This is due to the negative charge. Anyhow, positively charged particles will bend in opposite direction to that of electrons.

    Q.25: Why is it necessary to decrease the pressure in the discharge tube to get the cathode rays?

    Ans: The pressure in discharge tube is decreased to allow the cathode rays and anode rays to move freely from one electrons to other. In this way, the possibility of collisions between rays and the molecules are minimized.

    Q.26: Why the cathode rays are produced in the discharge tube by applying a high voltage?

    Ans: When the voltage of 5000-10000 volts is applied at the pressure of around 0.01 torr, then a glow appears and cathode rays travel from cathode towards the anode in the discharge tube. High voltage is necessary to break the molecules of the gas into atoms and to remove the electrons from outermost orbitals.

    Q.27: Why elm value of the cathode rays is just equal to that of electron?

    Ans: Since cathode rays are electrons, so their elm values are just equal to those of electrons.

    Q.28: Which observations tell the presence of cathode rays in the discharge tube?

    Ans: Cathode rays produced in the discharge tube fall upon the surface of the glass tube, just opposite to the cathode surface giving fluorescence. Fluorescence has different colours, which depends upon the nature of the glass tube.

    Q.29: Flow mass of au element is calculated from elm and charge?

    Ans: Mass of electron is calculated by using e/m value and charge of electron from Millikan oil chop experiment as:
    e/m = 1.7588 x 10¹¹ coulomb Kg”¹
    e = 1.6022 x 1019 coulombs

    Q.30: State Pauli’s exclusion iple and Hunds rule?

    Ans: According to Pauli principle no two electrons in an atom can have the same set of four quantum numbers. According to the Hund’s rule, if degenerate orbitals are available and more than two electrons are to be placed in them, then place in separate orbitals with the same spins rather than in the same orbital with the opposite spins.

    Q.31: Give the properties of neutron?

    Ans: PROPERTIES OF NEUTRONS:
    ➤ Neutron is electrically neutral particles.
    ➤ The mass of a neutron is almost the same as the mass of a proton.
    ➤ Mass of a neutron = 1.0087 a.m.u. or 1.6750 x 1027 kg.
    ➤ A free neutron decays into a proton with the emission of an electron and neutron.
    ➤ They can not ionize gases.
    ➤ They are highly penetrating particles.
    ➤ They can expel high speed protons from paraffin, water, paper and cellulose.
    ➤ Fast and slow neutrons.
    ➤ Neutrons are used as projectiles.
    ➤ When slow moving neutrons hit the Cu metal, radioactive Cu is produced along with the emission of Gamma Rays.

    Q.32: Give the properties of a, ß, y rays?

    Ans: PROPERTIES OF α RAYS:
    ➤ α – rays consist of a particle. Each a particle consists of ₂He nucleus.
    ➤ α- particle carry positive charge. Mass of each a particle is 4 times that o or H-atom.lonization power of a rays is very high.
    ➤ Penetration power of a rays is very small.
    ➤ α-rays produce fluorescence in different substances. a rays produce burn and source on human body.
    ➤ α-rays can produce artificial radioactivity in certain nuclei.
    ➤ They have strong ionization power because they remove electrons from the atoms of gas through which they pass. Blogspot.com
    ➤ Their velocity range is 3 x 10′ m/s to 3 x 10 m/s
    PROPERTIES OF ẞ RAYS:
    ➤ ẞ-rays consist of fast moving electrons,
    ➤ ẞ-rays have negative charge.
    ➤ Velocity of ẞ-rays is from 9 x 10′ m/sec to 27 x 10′ m/sec.
    ➤ ẞ-rays affect the photographic plate.
    ➤ lonization power of ẞ-rays is very small.
    ➤ Kinetic energy of ẞ-rays is less than that of a – rays.
    ➤ ẞ-rays produce fluorescence in different substance.
    PROPERTIES OF Y RAYS:
    ➤ y-rays are electromagnetic radiations.
    ➤ y-rays have no charge.
    ➤ y-rays travel with the velocity of light that is 3 x 108 m/sec.
    ➤ Penetration power of y-rays is very large. It is about hundred times larger than that of B-rays.
    ➤ y-rays produce feeble fluorescence when incident on screen coated with barium platino cyanide.

    Q.33: Give the properties of Cathode Rays?

    Ans: PROPERTIES OF CATHODE RAYS:
    ➤ These rays originate from cathode.
    ➤ Cathode rays travel in straight line.
    ➤ Cathode rays exert mechanical pressure.
    ➤ The cathode rays consist of material particles because they produce shadow of objects placed in the way.
    ➤ Cathode rays deflect in electric field towards the positive terminal.
    ➤ Cathode rays deflected by magnetic field.
    ➤ Cathode rays penetrate small thickness of matter such as alumimium foil, gold foil etc.
    ➤ These rays carry a negative charge.

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    the purpose of these notes was to make them as effective as possible. However, mistakes are still possible no matter how hard we try. In any case, if you see them, please let us know by commenting below. We appreciate your ideas and suggestions for improving the study material. Our efforts are meant to benefit all of the community, so we encourage you to share them with your friends, as “Sharing is caring“.