NCERT Solutions for Class 12 Physics Chapter 12 Atoms

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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.

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NCERT Solutions for Class 12 Physics Chapter 12 Atoms

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 solutions of NCERT 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

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

Q 12.1 Choose the correct alternative from the clues given at the end of the each statement:

(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.)

(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.)

Answer:

(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.

(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 .

Q 12.2 Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Answer:

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.

Q 12.3 What is the shortest wavelength present in the Paschen series of spectral lines?

Answer:

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.

Q 12.4 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Answer:

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

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

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

Q 12.5 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Answer:

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.

Q 12.6 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

Answer:

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.

Q 12.7 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels

Answer:

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

Q 12.7 (b) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels

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

Answer:

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

Q 12.8 The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10 ^{-11}m$ . What are the radii of the n = 2 and n =3 orbits?

Answer:

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

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

Q 12.9 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Answer:

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

Q 12.10 In accordance with the Bohr’s model, find 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 orbital speed $3 \times 10^{4}m/s$ (Mass of earth = $6.0 \times 10^{24}kg$ .)

Answer:

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

Q 12.11 (a) Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

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?

Answer:

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

Q 12.11 (b) Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

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?

Answer:

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

Q 12.11 (c) Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

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?

Answer:

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.

Q 12.11 (d) Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

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?

Answer:

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.

Q 12.12 The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about $10^{-40}$ . An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Answer:

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.

Q 12.13 Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Answer:

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.

Q 12.14 (a) Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not,say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom ( $\approx 10^{-10}m$ ).

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

Answer:

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.

Q 12.14 (b) Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom ( $\approx 10^{-10}m$ ).

(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.

Answer:

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.

Q 12.15 (a) The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.

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

Answer:

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

K=-E

K=-(-3.4)

K=3.4 eV

Q 12.15 (b) The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.

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

Answer:

Total Energy= Potential energy + Kinetic Energy

E=U+K

U=E-K

U=-3.4-3.4

U=-6.8 eV

Q 12.15 (c) The total energy of an electron in the first excited state of the hydrogen atom is about - 3.4eV.

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

Answer:

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

Q 12.16 If Bohr’s quantisation postulate (angular momentum = $\frac{nh}{2\pi }$ ) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?

Answer:

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.

Q 12.17 Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon ( $\mu^{-}$ ) of mass about 207m _e orbits around a proton].

Answer:

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

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NCERT solutions for class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

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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.

NCERT solutions subject wise

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

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.
Elaborate Explanations: Each class 12 atoms ncert solutions provides comprehensive, step-by-step explanations, making intricate ideas accessible to students.
Clear and Simple Language: The class 12 physics atoms ncert solutions are articulated in plain and straightforward language, guaranteeing easy comprehension.
Practice Opportunities: Class 12 physics ch 12 ncert solutions are included for practice and self-assessment, enhancing students' problem-solving skills.
Exam Readiness: These atoms physics class 12 ncert solutions are vital for board exam preparation and offer invaluable assistance for competitive exams.
Foundation for Advanced Studies: The concepts explored in this chapter establish the groundwork for more advanced studies in atomic physics and related fields.
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:

Frequently Asked Questions (FAQs)

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.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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.

Read Complete Answer

Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

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.

Read Complete Answer

kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

NCERT Solutions for Class 12 Physics Chapter 12 Atoms

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

NCERT Solutions for Class 12 Physics Chapter 12 Atoms

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

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