Atoms Chapter 12: NCERT Solutions for Class 12

Atoms Chapter 12 is an eminent part of Class 12th syllabus focusing on the vital theories including Rutherford’s model, Paschen’s series of spectral lines, etc. Students must take the chapter as a priority to score well in their Physics paper. 

SimplyAcad has provided best the solutions of Class 12 Atoms Chapter 12 for students to help them clear all the doubts and confusions regarding the notions of atoms. The solutions will make sure to prepare students for their upcoming boards examination, scroll below to find the answers organised in an structured manner for students’ ease.

 

Class 12 Physics NCERT Solutions Chapter 12 Atom Important Questions

Q1: Choose the correct alternative from the clues given at the end of 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.)

(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 ………. 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 models.)

Ans:

(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 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 models.

Q2: 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?

Ans:

We know that the mass of the incident alpha particle (6.64 × 10-27kg) is more than the mass of hydrogen (1.67 × 10-27Kg). Hence, the target nucleus is lighter, from which we can conclude that the alpha particle would not rebound, implying the fact that solid hydrogen isn’t a suitable replacement for gold foil for the alpha particle scattering experiment.

Q3: What is the shortest wavelength present in the Paschen series of spectral lines? \section*

{Solution} The shortest wavelength in the Paschen series corresponds to the transition from \( n = \infty \) to \( n’ = 3 \). The Rydberg formula for the wavelength of spectral lines is given by: \[ \frac{1}{\lambda} = R \left( \frac{1}{n’^2} – \frac{1}{n^2} \right) \] where \( R \) is the Rydberg constant, \( n’ \) is the lower energy level, and \( n \) is the higher energy level. For the Paschen series, \( n’ = 3 \) and the transition occurs from \( n = \infty \), so: \[ \frac{1}{\lambda} = R \left( \frac{1}{3^2} – \frac{1}{\infty^2} \right) \] Simplifying this: \[ \frac{1}{\lambda} = R \left( \frac{1}{9} \right) \] The Rydberg constant \( R \) is approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \). Therefore: \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{1}{9} = 1.219 \times 10^6 \, \text{m}^{-1} \] Taking the reciprocal to find the wavelength: \[ \lambda = \frac{1}{1.219 \times 10^6} \approx 8.20 \times 10^{-7} \, \text{m} = 820.4 \, \text{nm} \] \textbf{Final Answer:} The shortest wavelength present in the Paschen series is approximately \( 820.4 \, \text{nm} \). \end{document}

Q4: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Ans:

It is given that

Separation of two energy levels in an atom,

E = 2.3 eV = 2.3 × 1.6 × 10-19

= 3.68 × 10-19 J

Consider v as the frequency of radiation emitted when the atom transits from the upper level to the lower level.

So, the relation for energy can be written as:

E = hv

Where,

h = Planck’s constant = 6.62 × 10-34 Js

v = E/h

= 

= 5.6 × 1014 Hz

Therefore, the frequency of radiation is 5.6 × 1014 Hz.

Q5: The ground state energy of the hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Ans:

Ground state energy of hydrogen atom, E = − 13.6 eV

The total energy of the hydrogen atom is -13.6 eV. The kinetic energy is equal to the negative of the total energy.

Kinetic energy = − E = − (− 13.6) = 13.6 eV

Potential energy = negative of two times of kinetic energy.

Potential energy = −2 × ( 13.6 ) = −27.2 eV.

Q6: 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 the photon. \section*

{Solution} The energy levels of a hydrogen atom are given by: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] For the ground level (\( n = 1 \)): \[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \] For the level \( n = 4 \): \[ E_2 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, \text{eV} \] The energy absorbed by the photon is: \[ E = E_2 – E_1 = (-0.85 \, \text{eV}) – (-13.6 \, \text{eV}) = 13.6 – 0.85 = 12.75 \, \text{eV} \] Converting this energy to joules: \[ E = 12.75 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.04 \times 10^{-18} \, \text{J} \] The energy of a photon is also given by: \[ E = \frac{hc}{\lambda} \] where \( h \) is Planck’s constant and \( c \) is the speed of light. Rearranging to solve for \( \lambda \): \[ \lambda = \frac{hc}{E} \] Substitute the values: \[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{2.04 \times 10^{-18} \, \text{J}} \approx 9.7 \times 10^{-8} \, \text{m} = 97 \, \text{nm} \] To find the frequency (\( \nu \)): \[ \nu = \frac{c}{\lambda} \] Substitute the values: \[ \nu = \frac{3 \times 10^8 \, \text{m/s}}{9.7 \times 10^{-8} \, \text{m}} \approx 3.1 \times 10^{15} \, \text{Hz} \] \textbf{Final Answer:} The wavelength of the photon is approximately \( 97 \, \text{nm} \) and the frequency is approximately \( 3.1 \times 10^{15} \, \text{Hz} \).

Q.7: (a) Using 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.

Ans:

(a) Let v1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1= 1. For charge (e) of an electron, v1 is given by the relation,

For level n2= 2, we can write the relation for the corresponding orbital speed as:

For level n3= 3, we can write the relation for the corresponding orbital speed as:

Therefore, in a hydrogen atom, the speed of the electron at different levels, that is, n = 1, n = 2, and n = 3, is 2.18×106m/s, 1.09×106m/s and 7.27 × 105 m/s.

(b) Let T1 be the orbital period of the electron when it is in level n1= 1.

The orbital period is related to orbital speed as:

h = Planck’s constant = 6.62 × 10−34 J s

e = Charge on an electron = 1.6 × 10−19 C

ε0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2

m = Mass of an electron = 9.1 × 10−31 kg

𝑇1=◂/▸

For level n2 = 2, we can write the period as:

For level n3 = 3, we can write the period as:

Therefore, 1.52×10-16s, 1.22×10-15s and 4.12 × 10-15s are the orbital periods in each level.

Q8: The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n = 3 orbits?

Ans:

The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 × 10−11 m.

Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:

For n = 3, we can write the corresponding electron radius as:

Therefore, 2.12 × 10−10 m and 4.77 × 10−10 m are the radii of an electron for n = 2 and n

= 3 orbits, respectively.

Q9: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Ans:

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV, i.e., −1.1 eV.

Orbital energy is related to orbit level (n) as:

For n = 3, E = -13.6 / 9 = -1.5 eV

This energy is approximately equal to the energy of gaseous hydrogen.

It can be concluded that the electron has jumped from n = 1 to the n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for the Lyman series as:

Where, Ry= Rydberg constant = 1.097 × 107 m−1

λ= Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λ as:

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Therefore, there are two wavelengths that are emitted in the Lyman series, which are approximately 102.5 nm and 121.5 nm, and one wavelength in the Balmer series, which is 656.33 nm.

Q10: {Question} 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 the photon. \section*

{Solution} The energy levels of a hydrogen atom are given by: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] For the ground level (\( n = 1 \)): \[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \] For the level \( n = 4 \): \[ E_2 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, \text{eV} \] The energy absorbed by the photon is: \[ E = E_2 – E_1 = (-0.85 \, \text{eV}) – (-13.6 \, \text{eV}) = 13.6 – 0.85 = 12.75 \, \text{eV} \] Converting this energy to joules: \[ E = 12.75 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.04 \times 10^{-18} \, \text{J} \] The energy of a photon is also given by: \[ E = \frac{hc}{\lambda} \] where \( h \) is Planck’s constant and \( c \) is the speed of light. Rearranging to solve for \( \lambda \): \[ \lambda = \frac{hc}{E} \] Substitute the values: \[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{2.04 \times 10^{-18} \, \text{J}} \approx 9.7 \times 10^{-8} \, \text{m} = 97 \, \text{nm} \] To find the frequency (\( \nu \)): \[ \nu = \frac{c}{\lambda} \] Substitute the values: \[ \nu = \frac{3 \times 10^8 \, \text{m/s}}{9.7 \times 10^{-8} \, \text{m}} \approx 3.1 \times 10^{15} \, \text{Hz} \] \textbf{Final Answer:} The wavelength of the photon is approximately \( 97 \, \text{nm} \) and the frequency is approximately \( 3.1 \times 10^{15} \, \text{Hz} \). 

Additional Questions:

Q11: Choose a suitable solution to the given statements which justify the difference between Thomson’s model and Rutherford’s model

(a) In the case of scattering of alpha particles by a gold foil, the average angle of deflection of alpha particles stated by Rutherford’s model is (less than, almost the same as, or much greater than) stated by Thomson’s model.

(b) Is the likelihood of reverse scattering (i.e., dispersing of α-particles at points more prominent than 90°) anticipated by Thomson’s model  (considerably less, about the same, or much more prominent) than that anticipated by Rutherford’s model?

(c) For a small thickness T, keeping other factors constant, it has been found that the amount of alpha particles scattered at direct angles is proportional to T. What does this linear dependence imply?

(d) To calculate the average angle of scattering of alpha particles by thin gold foil, which model states that it’s wrong to skip multiple scattering?

Ans:

(a) almost the same

The normal point of diversion of alpha particles by a thin gold film anticipated by Thomson’s model is about the same as anticipated by Rutherford’s model. This is on the grounds that the average angle was taken in both models.

(b) considerably less

The likelihood of scattering of alpha particles at points more than 90° anticipated by Thomson’s model is considerably less than that anticipated by Rutherford’s model.

(c) Dispersing is predominantly because of single collisions. The odds of a single collision increment linearly with the amount of target molecules. Since the number of target particles increments with an expansion in thickness, the impact likelihood depends straightly on the thickness of the objective.

(d) Thomson’s model

It isn’t right to disregard multiple scattering in Thomson’s model for figuring out the average angle of scattering of alpha particles by a thin gold film. This is on the grounds that a solitary collision causes almost no deflection in this model. Subsequently, the watched normal scattering edge can be clarified just by considering multiple scattering.

Q12: The gravitational attraction between an electron and a proton in a hydrogen atom is weaker than the Coulomb attraction by a factor of about \(10^{-40}\). Estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction instead of Coulomb attraction. \section*

{Solution} \subsection*{Step 1: Radius of the First Bohr Orbit} The radius of the first Bohr orbit in a hydrogen atom, based on Coulomb’s law, is given by: \[ a_0 = \frac{4 \pi \epsilon_0 h^2}{m_e e^2} \] where: \begin{itemize} \item \( \epsilon_0 \) is the permittivity of free space, \item \( h \) is Planck’s constant, \item \( m_e \) is the mass of the electron, \item \( e \) is the charge of the electron. \end{itemize} \subsection*{Step 2: Coulomb Force} The Coulomb force of attraction between an electron and a proton is given by: \[ F_C = \frac{e^2}{4 \pi \epsilon_0 r^2} \] \subsection*{Step 3: Gravitational Force} The gravitational force of attraction between an electron and a proton is given by: \[ F_G = \frac{G m_p m_e}{r^2} \] where: \begin{itemize} \item \( G \) is the gravitational constant (\(6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\)), \item \( m_p \) is the mass of the proton, \item \( m_e \) is the mass of the electron. \end{itemize} \subsection*{Step 4: Equating Gravitational and Coulomb Forces} Setting the gravitational force equal to the Coulomb force: \[ \frac{G m_p m_e}{r^2} = \frac{e^2}{4 \pi \epsilon_0 r^2} \] Solving for \( \frac{e^2}{4 \pi \epsilon_0} \): \[ \frac{e^2}{4 \pi \epsilon_0} = G m_p m_e \] \subsection*{Step 5: Substitute into Bohr Radius Formula} Substitute \( \frac{e^2}{4 \pi \epsilon_0} \) from above into the Bohr radius formula: \[ a_0 = \frac{h^2}{4 \pi \epsilon_0 m_e e^2} \] Replacing \( \frac{e^2}{4 \pi \epsilon_0} \) with \( G m_p m_e \): \[ a_0 = \frac{h^2}{G m_p m_e} \] Substitute the known values: \[ h = 6.626 \times 10^{-34} \, \text{Js} \] \[ G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \] \[ m_p = 1.67 \times 10^{-27} \, \text{kg} \] \[ m_e = 9.1 \times 10^{-31} \, \text{kg} \] \[ a_0 = \frac{(6.626 \times 10^{-34})^2}{6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times (9.1 \times 10^{-31})} \] \[ a_0 \approx 1.21 \times 10^{29} \, \text{m} \] \subsection*{Conclusion} The radius of the first Bohr orbit if the electron and proton were bound by gravitational attraction is much greater than the estimated size of the observable universe, which is about \(1.5 \times 10^{27} \, \text{m}\). Thus, the gravitational attraction between the electron and proton is indeed extremely weak compared to the Coulomb attraction. \textbf{Final Answer:} The radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

Q13: {Question} 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. \section*

{Solution} \subsection*{1. Expression for Frequency of Radiation Emitted} The energy levels of a hydrogen atom are given by: \[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \] where \( n \) is the principal quantum number. When an electron transitions from level \( n \) to level \( n_1 \), the change in energy \( \Delta E \) is: \[ \Delta E = E_{n_1} – E_n \] Substituting the energy expressions: \[ \Delta E = -\frac{13.6 \, \text{eV}}{n_1^2} – \left(-\frac{13.6 \, \text{eV}}{n^2}\right) \] \[ \Delta E = 13.6 \, \text{eV} \left(\frac{1}{n_1^2} – \frac{1}{n^2}\right) \] The frequency of the emitted photon is given by: \[ \nu = \frac{\Delta E}{h} \] Substituting \( \Delta E \): \[ \nu = \frac{13.6 \, \text{eV}}{h} \left(\frac{1}{n_1^2} – \frac{1}{n^2}\right) \] \subsection*{2. Classical Frequency of Revolution for Large \( n \)} The radius of the \( n \)-th Bohr orbit is: \[ r_n = \frac{4 \pi \epsilon_0 \left(\frac{h}{2 \pi}\right)^2 m_e}{e^2 n^2} \] The velocity of the electron in the \( n \)-th orbit is: \[ v_n = \frac{e^2}{4 \pi \epsilon_0 h} \frac{1}{n} \] The classical frequency of revolution \( \nu_e \) is given by: \[ \nu_e = \frac{v_n}{2 \pi r_n} \] Substituting \( v_n \) and \( r_n \): \[ \nu_e = \frac{\frac{e^2}{4 \pi \epsilon_0 h} \frac{1}{n}}{2 \pi \frac{4 \pi \epsilon_0 \left(\frac{h}{2 \pi}\right)^2 m_e}{e^2 n^2}} \] Simplify: \[ \nu_e = \frac{e^4 n^2}{16 \pi^3 \epsilon_0^2 h^3 m_e} \] For large \( n \), the frequency of radiation \( \nu \) emitted during the transition can be approximated by: \[ \nu \approx \frac{13.6 \, \text{eV}}{h} \frac{1}{n^2} \] Thus, for large \( n \), this approximates to: \[ \nu \approx \frac{e^4}{16 \pi^3 \epsilon_0^2 h^3 m_e} \frac{1}{n^3} \] which matches the classical frequency of revolution of the electron in its orbit

Q 14: 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, a 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 textbook. 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 (~ 10–10m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.
(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 the energies of atoms are mostly in a non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr 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.

Ans:

(a)

Charge of an electron, e = 1.6 x 10-19 C

Mass of the electron, me = 9.1 x 10-31 kg

Speed of light, c = 3 x 108 m/s

The equation from the given quantities is given as,

ε0 is the permittivity of free space

The numerical value of the quantity is

= 9 x 109 x [ (1.6 x 10 -19)2/ 9.1 x 10-31 x (3 x 108)2]

= 2.81 x 10-15 m

The numerical value of the equation taken is much lesser than the size of the atom.

(b)

Charge of an electron, e = 1.6 x 10-19 C

Mass of the electron, me = 9.1 x 10-31 kg

Planck’s constant , h = 6.63 x 10-34  Js

Considering a quantity involving all these values as

Where, ε0 = Permittivity of free space

The numerical value of the above equation is

The numerical value of the quantity is the order of the atomic size.

Q 15: The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Ans:

(a) Total energy of the electron, E = -3.4 eV

The kinetic energy of the electron is equal to the negative of the total energy.

K.E = – E

= – (- 3.4 ) = + 3.4 eV

Kinetic energy =  + 3.4 eV

(b) Potential energy (U) of the electron is equal to the negative of twice the kinetic energy,

P.E = -2 (K.E)

= – 2 x 3.4 = -6.8 eV

Potential energy = -6.8 eV

(c) The potential energy of the system depends on the reference point. If the reference point is changed from zero, then the potential energy will change. The total energy is given by the sum of the potential energy and kinetic energy. Therefore, the total energy will also change.

 

Q16: If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion as well. Why, then, do we never speak of the quantisation of the orbits of planets around the Sun?

Ans:

The quantum level for a planetary motion is considered to be continuous. This is because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). 1070h is the order of the angular momentum of the Earth in its orbit. As the values of n increase, the angular momenta decreases. So, planetary motion is considered to be continuous.

Q17: 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 (μ−) of mass about 207me orbits around a proton].

Ans:

Mass of a negatively charged muon,  mμ = 207 me

According to Bohr’s model:

Bohr radius,

And, the energy of a ground state electronic hydrogen atom, Ee  

Also, the energy of a ground state muonic hydrogen atom, Eμ α mμ

We have the value of the first Bohr orbit, re  = 0.53 A = 0.53 × 10-10 m

Let rμ be the radius of the muonic hydrogen atom.

Following is the relation at equilibrium:

Hence, for a muonic hydrogen atom, 2.56 × 10−13 m is the value of the first Bohr radius.

We have,

Ee= − 13.6 eV

Considering the ratio of the energies, we get

Eμ= 207 Ee = 207 × ( – 13.6 ) = – 2.81 k eV

−2.81 k eV is the ground state energy of a muonic hydrogen atom.