NCERT Solutions for Class 12 Physics Chapter 12 Atoms

# NCERT Solutions for Class 12 Physics Chapter 12 Atoms

Edited By Vishal kumar | Updated on Sep 12, 2023 12:27 PM IST | #CBSE Class 12th

## NCERT Solutions for Class 12 Physics Chapter 12 – Free PDF Download

NCERT Solutions for Class 12 Physics Chapter 12 Atoms: Welcome to the updated atoms ncert solutions class 12. On this Careers360 page, you'll find detailed question-and-answer explanations crafted by subject experts. These class 12 atoms ncert solutions are available in PDF format for students' convenience, enabling access anytime and anywhere according to their preferences.

Do you know that an atom is electrically neutral, which tells that atoms contain an equal number of positive and negative charges. The Atoms Class 12 NCERT Solutions explain questions related to different models of the atom, their drawbacks, hydrogen spectra, and Debroglie hypothesis. CBSE NCERT solutions for Class 12 Physics chapter 12 atoms will help you in board exam preparation. The plays a major role in making the concepts easy and hence will help in competitive exams also. The questions in Atoms Class 12 NCERT text book are divided in to exercise and additional exercise. All questions are solved based on the NCERT syllabus.

Studying NCERT solutions for Class 12 Physics Chapter 12 Atoms is important because it provides a clear and detailed understanding of the questions covered in the chapter. It also helps in preparing for exams like JEE or NEET and understanding the fundamental principles of physics.

Free download class 12 physics atoms ncert solutions pdf for CBSE exam.

## NCERT Solutions for Class 12 Physics Chapter 12 Atoms

NCERT Solutions for Class 12 Physics Chapter 12 Atoms: Exercise Solution

(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)

(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)

(c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model.)

(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/ Rutherford’s model.)

(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models.)

(a) The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model.

(b) In the ground state of Thomson’s model electrons are in stable equilibrium, while in Rutherford’s model electrons always experience a net force.

(c) A classical atom based on Rutherford’s model is doomed to collapse.

(d) An atom has a nearly continuous mass distribution in a Thomson’s model but has a highly non-uniform mass distribution in Rutherford’s model .

(e) The positively charged part of the atom possesses most of the mass in both the models .

On repeating the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil we would have different observations than Rutherford, as the alpha particles won't be scattered much because of being heavier than the nucleus of the Hydrogen atom. Therefore we would not be able to confirm the presence of almost the entire mass of the atom at its centre.

The Rydberg's formula for the hydrogen atom is

$\frac{1}{\lambda }=R\left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ]$

Where R is Rydberg constant for the Hydrogen atom and equals to 1.1 $\times$ 10 7 m -1

For shortest wavelength in Paschen Series n 1 =2 and n 2 = $\infty$

$\frac{1}{\lambda }=1.1\times 10^{7}\left [ \frac{1}{3^{2}}-\frac{1}{\infty^{2}} \right ]$

$\lambda =8.18\times 10^{-7}\ m$

The shortest wavelength in Paschen Series is therefore 818 nm.

Frequency of radiation consisting of photons of energy E is given by

$\nu =\frac{E}{h}$

E=2.3 eV

Plank's constant(h)=6.62 $\times$ 10 -34 Js

$\nu =\frac{2.3\times 1.6\times 10^{-19}}{6.62\times 10^{-34}}$

$\nu =5.55\times 10^{14}\ Hz$

The ground state energy E=-13.6 eV.

The kinetic energy= -E=13.6 eV

Also ground state energy = Kinetic energy+Potential energy

E=K+U

U=E-K

U=-13.6-13.6

U=-27.2 eV

The kinetic and potential energies are 13.6 eV and -27.2 eV respectively.

The initial energy of the electron is E 1

$E_{1}=-\frac{13.6}{1^{2}}$

E 1 =-13.6 eV

The energy of the electron when it is excited to level n=4 is E 2

$E_{1}=-\frac{13.6}{4^{2}}$

E 2 =-0.85 eV

The difference between these two energy levels is equal to the energy of the photon absorbed by the electron.

The energy of the photon $\Delta$ E = E 2 - E 1

$\Delta$ E = -0.85 -(-13.6)

$\Delta$ E = 12.75 eV

The wavelength of the photon can be calculated using relation

$\Delta E=\frac{hc}{\lambda }$

hc=1240 eV

$\\\lambda =\frac{hc}{\Delta E}\\ \lambda=\frac{1240}{12.57}\\ \lambda=98.6\ nm$

$\\\nu =\frac{c}{\lambda }\\ \nu =\frac{3\times 10^{8}}{98.6\times 10^{-9}}\\\nu =3.04\times 10^{15}\ Hz$

The wavelength and frequency of the photon absorbed by the hydrogen atom are 98.6 nm and 3.04 $\times$ 10 15 Hz respectively.

As per Bohr's model the angular momentum of electrons in each orbit is constant and a multiple of $\frac{nh}{2\pi }$

$m_{e}v_{n}r_{n}=\frac{nh}{2\pi }$ (i)

The electrostatic force of attraction between the electron and the nucleus provides the required centripetal force for the circular motion of the electron.

$\frac{mv_{n}^{2}}{r_{n}}=\frac{e^{2}}{4\pi \varepsilon _{0}r_{n}^{2} }$ (ii)

Using equation (i) and (ii) we get

$v_{n}=\frac{e^{2}}{2nh\varepsilon _{_{0}}}$

$r_{n}=\frac{n^{2}h^{2}\varepsilon _{_{0}}}{m_{e}\pi e^{2}}$

$\\v_{1}=\frac{e^{2}}{2h\varepsilon _{_{0}}}\\ \\v_{1}=\frac{(1.6\times 10^{-19})^{2}}{2\times 6.62\times 10^{-34}\times 8.85\times 10^{-12}}$

v 1 =2.18 $\times$ 10 6 ms -1

$\\v_{2}=\frac{e^{2}}{4h\varepsilon _{_{0}}}\\ \\v_{1}=\frac{(1.6\times 10^{-19})^{2}}{4\times 6.62\times 10^{-34}\times 8.85\times 10^{-12}}$

v 2 =1.09 $\times$ 10 6 ms -1

$\\v_{3}=\frac{e^{2}}{6h\varepsilon _{_{0}}}\\ \\v_{1}=\frac{(1.6\times 10^{-19})^{2}}{6\times 6.62\times 10^{-34}\times 8.85\times 10^{-12}}$

v 3 =7.28 $\times$ 10 5 ms -1

(b) calculate the orbital period in each of these levels.

Orbital period (T n ) is defined as time taken by the electron to complete one revolution around the nucleus and is given by

$\\T_{n}=\frac{2\pi r_{n}}{v_{n}}\\ T_{n}=\frac{4n^{3}h^{3}\varepsilon _{0}^{2}}{m_{e}e^{4}}$

$T_{1}=\frac{4\times 1^{3}\times (6.62\times 10^{-34})^{2} \times (8.85\times 10^{-12})^{2}}{9.1\times 10^{-31}\times (1.6\times 10^{-19})^{4}}$

T 1 =1.53 $\times$ 10 -16 s

$T_{2}=\frac{4\times 2^{3}\times (6.62\times 10^{-34})^{2} \times (8.85\times 10^{-12})^{2}}{9.1\times 10^{-31}\times (1.6\times 10^{-19})^{4}}$

T 2 =1.22 $\times$ 10 -15 s

$T_{3}=\frac{4\times 3^{3}\times (6.62\times 10^{-34})^{2} \times (8.85\times 10^{-12})^{2}}{9.1\times 10^{-31}\times (1.6\times 10^{-19})^{4}}$

T3=4.12 $\times$ 10 -15 s

The radius of the orbit is proportional to the square of n.

For n=2 the radius of the orbit is

$\\r_{2}=r_{1}\times 2^{2}\\ =5.3\times 10^{-11}\times 4\\ =2.12\times 10^{-10}\ m$

For n=3 the radius of the orbit is

$\\r_{3}=r_{1}\times 3^{2}\\ =5.3\times 10^{-11}\times 9\\ =4.77\times 10^{-10}\ m$

Since the energy of the electron beam is 12.5 eV the Hydrogen atoms will get excited to all requiring energy equal to or less than 12.5 eV

E 1 =-13.6 eV

E 3 = -1.5 eV

E 3 -E 1 = 12.1 eV

E 4 = -0.85 eV

E 4 -E 1 =12.75 eV

Therefore the electron can reach maximum upto the level n=3.

During de-excitations, the electron can jump directly from n=3 to n=1 or it can first jump from n=3 to n=2 and then from n=2 to n=1

Therefore two wavelengths from the Lyman series and one from the Balmer series will be emitted

To find the wavelengths emitted we will use the Rydberg's Formula

$\frac{1}{\lambda }=R(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$ where R is the Rydberg's constant and equals 1.097 $\times$ 10 7 m -1

For n 1 =1 and n 2 =3

$\frac{1}{\lambda }=1.097\times 10^{7}(\frac{1}{1^{2}}-\frac{1}{3^{2}})$

Emitted wavelength is 102.5 nm

For n 1 =1 and n 2 =2

$\frac{1}{\lambda }=1.097\times 10^{7}(\frac{1}{1^{2}}-\frac{1}{2^{2}})$

Emitted wavelength is 121.54 nm

For n 1 =2 and n 2 =3

$\frac{1}{\lambda }=1.097\times 10^{7}(\frac{1}{2^{2}}-\frac{1}{3^{2}})$

Emitted wavelength is 656.3 nm

As per the Bohr's model, the angular of the Earth will be quantized and will be a multiple of $\frac{h}{2 \pi}$

$\\mvr=\frac{nh}{2 \pi}\\ n=\frac{2 \pi mvr}{h}\\ n=\frac{2\pi\times 6\times 10^{24}\times 3\times 10^{4}\times 1.5\times 10^{11}}{6.62\times 10^{-34}}$

n = 2.56 $\times$ 10 74

Therefore the quantum number that characterises the earth’s revolution around the sun in an orbit of radius $1.5 \times 10^{11}m$ m with an orbital speed $3 \times 10^{4}m/s$

is 2.56 $\times$ 10 74

## NCERT solutions for class 12 physics chapter 12 atoms: Additional Exercise Solution

Is the average angle of deflection of $\alpha$ -particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

The average angle of deflection of $\alpha$ -particles by a thin gold foil predicted by both the models is about the same.

Is the probability of backward scattering (i.e., scattering of $\alpha$ -particles at angles greater than $90^{\circ}$ ) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

The probability of backward scattering predicted by Thomson’s model is much less than that predicted by Rutherford’s model.

Keeping other factors fixed, it is found experimentally that for small thickness t, the number of $\alpha$ -particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

Scattering at moderate angles requires head-on collision the probability of which increases with the number of target atoms in the path of $\alpha$ -particles which increases linearly with the thickness of the gold foil and therefore the linear dependence between the number of $\alpha$ -particles scattered at a moderate angle and the thickness t of the gold foil.

In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of $\alpha$ -particles by a thin foil?

It is completely wrong to ignore multiple scattering for the calculation of the average angle of scattering of $\alpha$ -particles by a thin foil in Thomson's model as the deflection caused by a single collision in this model is very small.

As per the bohrs model

$m_{e}v_{n}r_{n}=\frac{nh}{2\pi}$ (i)

If the proton and the electron were bound only by the gravity the gravitational force between them will provide the centripetal force required for circular motion

$\frac{m_{e}v_{n}^{2}}{r_{n}}=\frac{Gm_{e}m_{p}}{r_{n}^{2}}$ (ii)

From equation (i) and (ii) we can calculate that the radius of the ground state (for n=1) will be

$\\r_{1}=\frac{h^{2}}{4\pi Gm_{p}m_{e}^{2}}\\ r_{1}=\frac{(6.62\times 10^{-34})^{2}}{4\pi \times 6.67\times 10^{-11}\times 1.67\times 10^{-27}\times (9.1\times 10^{-31})^{2}}$

$r_{1}\approx 1.2\times 10^{29}\ m$

The above value is larger in order than the diameter of the observable universe. This shows how much weak the gravitational forces of attraction as compared to electrostatic forces.

Using Bohr's model we have.

$v_{n}=\frac{e^{2}}{2nh\varepsilon _{_{0}}}$

$r_{n}=\frac{n^{2}h^{2}\varepsilon _{_{0}}}{m_{e}\pi e^{2}}$

$\\E_{n}=\frac{1}{2}mv_{n}^{2}-\frac{e^{2}}{4\pi \epsilon _{0}r_{n}^{2}}\\ \\E_{n}=-\frac{me^{4}}{8n^{2}h^{2} \epsilon_{0}^{2} }$

$\\E_{n}-E_{n-1}=-\frac{me^{4}}{8n^{2}h^{2} \epsilon_{0}^{2} }-(-\frac{me^{4}}{8(n-1)^{2}h^{2} \epsilon_{0}^{2} })\\ \\E_{n}-E_{n-1}=-\frac{me^{4}}{8h^{2} \epsilon_{0}^{2} }[\frac{1}{n^{2}}-\frac{1}{(n-1)^{2}}]\\E_{n}-E_{n-1}=-\frac{me^{4}}{8h^{2} \epsilon_{0}^{2} }[\frac{-2n+1}{n^{2}(n-1)^{2}}]$

Since n is very large 2n-1 can be taken as 2n and n-1 as n

$\\E_{n}-E_{n-1}=-\frac{me^{4}}{8h^{2} \epsilon_{0}^{2} }[\frac{-2n}{n^{2}(n)^{2}}]\\ \\E_{n}-E_{n-1}=\frac{me^{4}}{4n^{3}h^{2} \epsilon_{0}^{2} }$

The frequency of the emission caused by de-excitation from n to n-1 would be

$\\\nu =\frac{E_{n}-E_{n-1}}{h}\\ \nu =\frac{me^{4}}{4n^{3}h^{3} \epsilon_{0}^{2} }$

The classical frequency of revolution of the electron in the nth orbit is given by

$\nu =\frac{v_{n}}{2\pi r_{n}}$

$\nu =\frac{e^{2}}{2nh\epsilon _{_{0}}}\times \frac{m_{e}\pi e^{2}}{2\pi n^{2}h^{2}\epsilon _{_{0}}}$

$\nu =\frac{me^{4}}{4n^{3}h^{3} \epsilon_{0}^{2} }$

The above is the same as the frequency of the emission during de-excitation from n to n-1.

(a) construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.

Using dimensional analysis we can see that the quantity to be constructed and consisting of m e , e and c will also have $\epsilon _{0}$ and will be equal to

$\frac{e^{2}}{\epsilon _{0}m_{e}c^{2}}$ and has numerical value 3.5 $\times$ 10 -14 which is much smaller than the order of atomic radii.

(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.

Using dimensional analysis we can see that the quantity to be constructed and consisting of m e , e and h will also have $\epsilon _{0}$ and will be equal to

$\frac{\epsilon _{0}h^{2}}{m_{e}e^{2}}$ and has a numerical value of approximately 6.657 $\times$ 10 -10 which is about the order of atomic radii.

What is the kinetic energy of the electron in this state?

Since we know that kinetic energy is equal to the negative of the total energy

K=-E

K=-(-3.4)

K=3.4 eV

What is the potential energy of the electron in this state?

Total Energy= Potential energy + Kinetic Energy

E=U+K

U=E-K

U=-3.4-3.4

U=-6.8 eV

Which of the answers above would change if the choice of the zero of potential energy is changed?

The total energy would change if the choice of the zero of potential energy is changed.

We never speak of Bohr's quantization postulate while studying planetary motion or even motion of other macroscopic objects because they have angular momentum very large relative to the value of h. In fact, their angular momentum is so large as compared to the value of h that the angular momentum of the earth has a quantum number of order 10 70 . Therefore the angular momentum of such large objects is taken to be continuous rather than quantized.

As per Bohr's quantization postulate

$m_{\mu ^{-}}v_{n}r_{n}=\frac{nh}{2\pi }$

Similarly, like the case in a simple hydrogen atom, the electrostatic force acts centripetally

$\frac{m_{\mu ^{-}}v_{n}^{2}}{r_{n}}=\frac{e^{2}}{4\pi\epsilon _{0}r_{n}^{2} }$

From the above relations, we can see that in Bohr's model the Radius is inversely proportional to the mass of the orbiting body and Energy is directly proportional to the mass of the orbiting body.

In case of hydrogen, atom r 1 is 5.3 $\times$ 10 -11 m

Therefore in case of a muonic hydrogen atom

$r_{1}=\frac{5.3\times 10^{-11}}{207}$

r 1 = 2.56 $\times$ 10 -13 m

In case of the hydrogen atom, E 1 is -13.6 eV

Therefore in case of a muonic hydrogen atom

E 1 =207 $\times$ (-13.6)

E 1 =2.81 keV

## NCERT solutions for class 12 physics chapter wise

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields NCERT solutions for class 12 physics chapter 2 Electrostatic Potential and Capacitance NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism NCERT solutions for class 12 physics chapter 5 Magnetism and Matter NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current NCERT solutions for class 12 physics chapter8 Electromagnetic Waves NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions NCERT solutions for class 12 physics chapter 11 Dual nature of radiation and matter NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei NCERT solutions for class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

### Atoms Chapter Class 12 NCERT Solutions: Important Formulas and Diagrams

Atoms Class 12 comes under the unit Modern Physics. There are a seventeen questions discussed in the NCERT solutions for Class 12 Physics Class 12 Chapter 12 Atoms. Some of the important formulas of the chapter atoms which will help in NCERT solutions for class 12 physics are given below.

• According to Bohr’s postulates for stationary orbit angular momentum
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$L=\frac{nh}{2\pi}$

Where n is the quantum number

• The total energy of the electron in the stationary states of the hydrogen atom in electronvolt is given by

$E_n=-\frac{13.6}{n^2}eV$

• Another important formula in NCERT is the

De Broglie wavelength

$\lambda=\frac{h}{mv}$

Where ‘h’ is Planck's constant and mv is the momentum. This relation can be modified in terms of the kinetic energy of the particle. Try to derive it yourself.

## Importance of NCERT solutions for class 12 physics chapter 12 atoms:

For CBSE board exam on an average 3 to 4 marks questions are asked from the chapter Atoms. The solutions of NCERT Class 12 Physics chapter 12 Atoms will help in securing full marks in board exam for this chapter. As far as the NEET exam is considered up to 2 questions are expected from the chapter Atoms Class 12 Physics. The NCERT solutions for Class 12 Physics chapter 12 atoms will also help to perform better in exams like JEE Main and other competitive exams.

## Some of the important topics covered in Class 12 Physics Chapter 12 Atoms are:

• De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation

• Alpha-Particle Scattering and Rutherford’s Nuclear Model of Atom

• Alpha-particle Trajectory

• X-rays

• Bohr Model of the Hydrogen Atom

• The Line Spectra of the Hydrogen Atom

These topics are the main focus of the NCERT textbook and are likely to be given more emphasis in exams.

## Key Features of NCERT Solutions Class 12 Physics Atoms

1. In-Depth Coverage: These atoms ncert solutions class 12 comprehensively address all the topics and questions found in Chapter 12, ensuring a thorough grasp of atomic physics concepts.

2. Elaborate Explanations: Each class 12 atoms ncert solutions provides comprehensive, step-by-step explanations, making intricate ideas accessible to students.

3. Clear and Simple Language: The class 12 physics atoms ncert solutions are articulated in plain and straightforward language, guaranteeing easy comprehension.

4. Practice Opportunities: Class 12 physics ch 12 ncert solutions are included for practice and self-assessment, enhancing students' problem-solving skills.

5. Exam Readiness: These atoms physics class 12 ncert solutions are vital for board exam preparation and offer invaluable assistance for competitive exams.

6. Foundation for Advanced Studies: The concepts explored in this chapter establish the groundwork for more advanced studies in atomic physics and related fields.

7. Open Access: These atoms chapter class 12 ncert solutions are freely accessible, ensuring accessibility for all students.

These distinguishing features render Class 12 Physics Chapter 12 NCERT solutions an indispensable resource for students, facilitating their success in examinations and future academic endeavors.

### Also Check NCERT Books and NCERT Syllabus here:

1. How many questions from the chapter atoms comes in CBSE board exams?

One or two questions are usually asked in the CBSE board exams from Class 12 chapter Atoms. It can be a 2 or 3 mark question. Students should cover all the concepts of the Class 12 chapter Atoms. The question may be theoretical, numerical or derivations. The syllabus of CBSE board is based on the NCERT book. Students can practice questions based on NCERT syllabus for the board exam.

2. What is the importance of the chapter for engineering and medical entrance exams?

The Class 12 Physics chapter Atoms are important for both engineering and medical exams. At least one question can be expected from Atoms for both JEE Main and NEET.

3. What are the important topics of the Class 12 NCERT chapter Atoms

The Hydrogen spectrum, Energy of atoms, Bhor Model of Atom, Bhor Postulates etc are important topics of the Atoms Class 12 Physics Chapter. More questions related to these topics an be practiced from NCERT Exemplar for Class 12 Physics.

4. How class 12 physics atom is important for NEET?

class 12 physics atom is important for NEET because it forms the foundation of many important concepts in chemistry and physics, including atomic structure, quantum mechanics, wave-particle duality, chemical bonding, and the properties of chemical compounds.

5. What do you mean by according to class 12 physics chapter 12 ncert solutions?

Atomic spectra refer to the unique set of lines or bands emitted or absorbed by an atom when it is excited, they are used to identify elements and study the properties of atoms and their electronic structure. The study of atomic spectra is known as spectroscopy.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

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Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9