Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Atoms

1. The mass of a H-atom is less than the sum of the masses of a proton and electron. Why is this?

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Explanation- the reason behind this is the binding energy or we can say the mass defect.

BE= mass defect (931MeV ). So whatever energy is liberated in fission and fusion changes the mass slightly. That's why there is a defect in mass

2. Imagine removing one electron from He⁴ and He³. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.

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Explanation- On removing one electron from He4 and He3, the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass.Also after removing one electron from He4 and He3 atoms contain one electron and are hydrogen like atoms.

3. When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy? .

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Explanation- The transition of an electron from a higher energy to a lower energy level can appears in the form of electromagnetic radiation because electrons interact only electromagnetically.

4. Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+ 4/3)e and electron a charge (-3/4)e, where e = 1.6 x 10^-19 Give reasons for your answer.

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Explanation- If proton had a charge (+4/3)e and electron a charge (-3/4)e, then the Bohr formula for the H-atom remain same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.

5. Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?

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Explanation- According to Bohr model electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as

L=nh/2 $\pi L\propto$ n

6. Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?

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Explanation- The total energy of the electron in the stationary states of the hydrogen atom is given by E_n = - $\frac{m{e}^4}{8n^2{\epsilon }_0^2{h}^2}$

where signs are as usual and the m that occurs in the Bohr formula is the reduced mass of

electron and proton. Also, the total energy of the electron in the ground state of the

hydrogen atom is−13.6 eV. For H-atom reduced mass m_e . Whereas for positronium, the

reduced mass is m=m_e/2

Hence, the total energy of the electron in the ground state of the positronium atom is

 -13.6/2=-6.8eV

7. Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom.

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Explanation- For a He -nucleus with charge 2e and electrons of charge− e, the energy level in ground state is

E=-13.6Z²/n²= -54.4eV

Thus, the ground state will have two electrons each of energy E and the total ground state

energy would be   -(4×13.6)eV = -54.4eV.

8. Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

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Explanation- The electron in Hydrogen atom in ground state revolves on a circular path whose radius is equal to the Bohr radius (a_n) . Let the velocity of electron is v

∴ Number of revolutions per unit time= $\frac{2\pi a_0}{V}$

The electric current is given by i=q/t , if q charge flows in time t. Here,  q=e

The electric current is given by i= $\frac{2\pi a_0}{V}e$

9. Show that the first few frequencies of light that is emitted when electrons fall to nth level from levels higher than n, are approximate harmonics (i.e., in the ratio 1:2:3 …) when n >> 1.

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Explanation- The frequency of any line in a series in the spectrum of hydrogen like atoms corresponding to the transition of electrons from (n+ p) level to nth level can be expressed as a difference of two terms;

From the formula we can say that v=cRZ²{ $\frac{1}{({n+p)}^2}-\frac{1}{n^2}$ }

Where p =1,2,3,… R os rydberg constant

When p<

V=cRZ²{ $\frac{1}{n^2}({1+\frac{p}{n})}^{-2}-\frac{1}{n^2}$ }

V=cRZ²{ $\frac{1}{n^2}-\frac{2p}{n^3}-\frac{1}{n^2}\}$

By binomial theorem

V=cRZ²{ $\frac{2p}{n^3}=\frac{2cR{Z}^2}{n^3}p$

Thus, the first few frequencies of light that is emitted when electrons fall to the nth level

from levels higher than n , are approximate harmonic (i.e., in the ratio 1 : 2 : 3 …) when

n >> 1 .

10. What is the minimum energy that must be given to a H-atom in ground state so that it can emit an H_ґLine in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such H_ґ photon?

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Explanation- Hγ in Balmer series corresponds to transition n= 5 to n= 2. So, the electron in ground state i.e., from n= 1must first be placed in state n= 5.

Energy required for the transition from n= 2 to n= 5 is given by E₁-E₅=13.6-0.54=13.06eV

Since, angular momentum is conserved,

angular momentum corresponding to Hg photon= change in angular momentum of electron

so L₅-L₂=5h-2h= 3h = 3 $\times 1.06\times {10}^{-34}$

=3.18 $\times {10}^{-34}$ kgm²/s

1. The first four spectral in the Lyman series of a H-atom are λ= 1218 Å, 1028 Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider deuterium, calculate the shift in the wavelength of these lines.

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Explanation- as we know total energy in stationary orbit is

E_n=- $\frac{{-me}^4}{8n^2{{\epsilon }_0}^2{h}^2}$  where sign have usual meaning.

According to bohr third postulate h $\nu =E_{f-}{E}_i$

$\nu =$ - $\frac{{me}^4}{8{{\epsilon }_0}^2{h}^3}(\frac{1}{n_f^1}-\frac{1}{n_i^2})$

$\lambda \propto \frac{1}{\propto }$

Where $\propto$  is the reduced mass

Reduced mass for H=_H=;m_e(1-m_e/M)

D= _D; m_e(1-m_e/2M)

 =m_e(1-m_e/2M)(1+m_e/2M)

If for hydrogen deuterium, the wavelength

$\frac{{\lambda }_D}{{\lambda }_H}=\frac{{\propto }_H}{{\propto }_D}$ = (1+ $\frac{m_e}{2M}$ )^-1= (1- $\frac{1}{2\times 1840}$ )

${\lambda }_D={\lambda }_H\times 0.99973$  so lines emitted are 1217.7A⁰,1027.7A⁰,974.04A⁰

2. Deutrium was discovered in 1931by Harold Urey by measuring the small change in wavelength for a particular transition in ¹H and ² This is because, the wavelength of transition depends to a certain extent on the nuclear mass. If nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass μ , revolving around the nucleus at a distance equal to the electron-nucleus separation. Here μ = m_eM / ( m_e + M ) where M is the nuclear mass and m_e is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in . ¹H and ²H . (Mass of ¹H nucleus is 1.6725 × 10^-27kg , mass of  ¹H nucleus is 3.3374 × 10^-27kg 3.3374 × 10^-27 𝑘g , Mass of electron = 9.109 × 10^-31 kg ).

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Explanation- total energy of electron

E= $\frac{\mu Z^2{e}^4}{8{\epsilon }_o^2{h}^2}\frac{1}{n^2}$

The frequency hv= $\frac{\mu e^4}{8{\epsilon }_o^2{h}^2}\left(1-\frac{1}{4}\right)=\frac{\mu e^4}{8{\epsilon }_o^2{h}^2}\frac{3}{4}$

$∆\lambda$ = ${\lambda }_D-{\lambda }_H$

$100\times \frac{∆\lambda }{{\lambda }_H}$ = $\frac{{\lambda }_D-{\lambda }_H}{{\lambda }_H}\times 100$ = $\frac{{\mu }_D-{\mu }_H}{{\mu }_H}\times 100$

= $\frac{\frac{m_{e{M}_D}}{m_e+M_H}-\frac{m_{e{M}_D}}{m_e-M_H}}{\frac{m_{e{M}_D}}{m_e+M_H}}\times 100$

When m_e< H< D

after solving we get 2.714 $\times {10}^{-2}\%$

3. If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1 Å and (ii) R = 10  Å .

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Explanation- in this type of situation centripetal force is balanced by electrostatic force

So mv²/r= -ke²/r²

According to bohr postuates

Mvr = h when n=1

On solving n  $\frac{h^2}{m^2{r}^2}\frac{1}{r}={ke}^2/r^2$

So after solving r= 0.51A⁰

So potential energy  $\frac{{ke}^2}{r}$ = -27.2eV , KE= mv²/2

= $\frac{1}{2}m\frac{h^2}{r^2}=\frac{h}{2m{r}^2}=+13.6eV$

But if R

If R>>r the electron moves inside the sphere with radius r’

Charge inside r’⁴=e $\frac{{r\text{'}}^3}{R^3}$

So r’ = $\frac{h^2}{m}\frac{k}{e^2}\frac{R^3}{{r\text{'}}^3}$

So r’⁴= 0.51A⁰R³

= 510(A⁰)⁴

4. In the Auger process, an atom maizes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from a n – 2 to n = 1 transition.

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Explanation- the energy of nth state E_n=-Z²R $\frac{1}{n^2}$  where R is constant and Z=24

The energy release ina transition from 2 to 1 $∆E=Z^{2}R\left(1-\frac{1}{4}\right)=\frac{3}{4}{Z}^{2}R$

The energy required to eject a n=4 electron is E₄= Z²R1/16

So kinetic energy of auger electron is , KE= Z²R(3/4-1/16)=1/16Z²R

= $\frac{11}{16}\times 24\times 24\times 13.6eV=5385.6eV$

5. The inverse square law in electrostatics is F= $\frac{{\mathit{e}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}^2}$ for the force between an electron and proton . the (1/r) dependence of F can be understood in quantum theory as being due to fact the particles of light photon is massless. If photon had a mass m_p. force would be modified to F= $\frac{{\mathit{e}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}^2}\left[\frac{1}{{\mathit{r}}^2}+\frac{\mathit{\lambda }}{\mathit{r}}\right]\mathbf{e}\mathbf{x}\mathbf{p}(-\mathit{\lambda }\mathit{r})$  where $\mathit{\lambda }=\frac{{\mathit{m}}_{\mathit{p}}\mathit{c}\mathit{}}{\mathit{h}}$  and n= h/2 $\mathit{\pi }.$ estimate the change in the ground state energy of H atom if m_p were 10^-6 times the mass of electron.

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Explanation- m_p= 10^-6

M_pc²=10^{-6 $\times electronmass\times c^2$}

 = 0.8 $\times {10}^{-19}J$

Wavelength associated with both of them is same

$\frac{h}{m_{p}c}=\frac{hc}{m_p{c}^2}=\frac{{10}^{-34}\times 3\times {10}^8}{0.8\times {10}^{-19}}=4\times {10}^{-7}m$

U(r)=- $\frac{e^2}{4\pi {\epsilon }_o}\frac{\mathrm{e}\mathrm{x}\mathrm{p}(-\lambda t)}{r}$

Mvr=h

V=h/mr

Mv²/r=( $\frac{e^2}{4\pi {\epsilon }_o}$ ) ( $\frac{1}{r^2}+\frac{\lambda }{r}$ )

$\frac{h^2}{m{r}^3}=\left(\frac{e^2}{4\pi {\epsilon }_o}\right)\left(\frac{1}{r^2}-\frac{\lambda }{r}\right)$

$\frac{h^2}{m}=\frac{e^2}{4\pi {\epsilon }_o}{r}_B$

$kinecticwillbedoubleofgroundstateso$

13.6 $\times 2=27.2eV$

6. The Bohr model for the H-atom relies on the Coulomb ’s law of electrostatics. Coulomb’s law has not directly been verified for very
short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge +q₁, – q₂modified to

F= $\frac{{\mathit{k}\mathit{q}}_1{\mathit{q}}_2}{{\mathit{r}}^2}$ r>R₀

  = $\frac{{\mathit{k}\mathit{q}}_1{\mathit{q}}_2}{{\mathit{R}}_0^2}$ ( ${\frac{{\mathit{R}}_{\mathit{o}}}{\mathit{r}})}^{\mathit{\epsilon }}$  r 0

Calculate in such vase the ground state energy of a H atom , if E= 0.1 R₀= 1A⁰

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Explanation- when r 0=1A ^o

Let $\epsilon =2+\delta$

F= $\frac{{kq}_1{q}_2}{r^{2+\delta }}{R}_0^{\delta }$

Where $\frac{{kq}_1{q}_2}{r^2}$ = ( ${1.6\times {10}^{-19})}^2\times 9\times {10}^9=23.04\times {10}^{-29}N/m2$

Let p= $23.04\times {10}^{-29}N/m2$

Electrostatic force is balanced by centripetal force

Mv²/r or v²= $\frac{p{R}_0^{\delta }}{m{r}^{1+\delta }}$

1. Taking the Bohr radius as a₀= 53 pm, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr’s model, will be about
(a) 53 pm    

(b) 27 pm  

(c) 18 pm  

(d) 13 pm

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Answer-(c)

Explanation- The atomic number of lithium is 3, therefore, the radius of Li+ + ion in its ground state, on the basis of Bohr’s model, will be about 1/3 times to that of Bohr radius.

Therefore, the radius of lithium ion is near 53/3=18≈ pm.

2. The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is B= .

If one decides to work in a frame of reference where the electron is at rest,the proton would be moving around it.by similar arguments, the binding energy would be

B= (M=proton mass)

This last expression is not correct.

(a) N would not be integral

(b) Bohr-quantisation applies only two electron

(c) The frame in which the electron is at rest is not inertial

(d) The motion of the proton would not be in circular orbits, even approximately.

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Answer- c

Explanation- when one decides to work in a frame of reference where the electron is at rest, the given expression is not true as it the non-inertial frame of reference.

3. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with man electrons. This is because
(a) Of the electrons not being subject to a central force
(b) Of the electrons colliding with each other
(c) Of screening effects
(d) The force between the nucleus and an electron will no longer be given by Coulomb’s law

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Answer-(a)

Explanation-The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because of the electrons not being subject to a central force.

4. For the ground state, the electron in the H-atom has an angular .momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,
(a) Because Bohr model gives incorrect values of angular momentum
(b) Because only one of these Would Have a minimum energy
(c) Angular momentum must be in the direction of spin of electron
(d) Because electrons go around only in horizontal orbits

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Answer –(a)

Explanation- In the simple Bohr model, only the magnitude of angular momentum is kept equal to some integral multiple of h/2 π, where, h is Planck’s constant and thus, the Bohr model gives incorrect values of angular momentum.

5. 0₂ molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms
(a) Is not important because nuclear forces are short-ranged
(b) Is as important as electrostatic force for binding the two atoms
(c) Cancels the repulsive electrostatic force between the nuclei
(d) Is not important because oxygen nucleus have equal number of neutrons and protons

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Answer-(a)

Explantion- in the molecules, nuclear force between the nuclei of the two atoms is not important because nuclear forces are short-ranged.

6. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is
(a) 10.20 eV  

(b) 20.40eV    

(c) 13.6 eV

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Answer-(a)

Explanation -Total energy in ground state collide elastically Is 213.6eV= 27.2eV

When a electron excite from ground state to first excited state then

Energy is =-13.6/2²= 3.4eV+13.6(ground state energy) eV = 17.0 eV

So net energy is 27.2-17 =10.2eV

7. A set of atoms in an excited state decays
(a) In general to any of the states with lower energy
(b) Into a lower state only when excited by an external electric field
(c) All together simultaneously into a lower state
(d) To emit photons only when they collide .

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Answer-(a)

Explanation-A set of atoms in an excited state decays in general to any of the states with lower energy.

8. An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state,
(a) The electron would not move in circular orbits
(b) The energy would be (2)⁴times that of a H-atom
(c) The electrons, orbit would go around the protons
(d) The molecule will soon decay in a proton and a H-atom

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Answer- (a,c)

Explantion- The protons are separated by a small distance of the order of angstrom. In the ground state the electron would not move in circular orbits the electrons, orbit would go around the protons.

9. Consider aiming a beam of free electrons towards free protons. When-they scatter, an electron and a proton cannot combine to produce a H-atom.
(a) Because of energy conservation
(b) Without simultaneously releasing energy in the form of radiation
(c) Because of momentum conservation
(d) Because of angular momentum conservation

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Answer-(a,b)

Explantion- When beam of free electrons is aiming towards free protons. Then, they scatter but an electron and a proton cannot combine to produce a H-atom because of energy conservation and without simultaneously releasing energy in the form of radiation.

10. The Bohr model for the spectra of a H-atom
(a) Will not be applicable to hydrogen in the molecular form
(b) Will not be applicable as it is for a He-atom
(c) Is valid only at room temperature
(d) Predicts continuous as well as discrete spectral lines

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Answer-(a,b)

Explantion- The Bohr model for the spectra of a H-atom will not be applicable to hydrogen in the Molecular form. And also, it will not be applicable as it is for a He-atom.

11. The Balmer series for the H-atom can be observed

(a) If we measure the frequencies of light emitted when an excited atom falls to the ground state

(b) If we measure the frequencies of light emitted due to transitions between excited states and the first excited state

(c) In any transition in a H-atom .

(d) As a sequence of frequencies with the higher frequencies getting closely

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Answer- (b,d)

Explantion- Balmer series for the H-atom can be observed if we measure the frequencies of light emitted due to transitions between higher excited states and the first excited state and as a sequence of frequencies with the higher frequencies getting closely packed.

12. Let E_n= $\frac{-1\mathit{m}{\mathit{e}}^4}{8{\mathit{\epsilon }}_0^2{\mathit{n}}^2{\mathit{h}}^2}$ be the energy of the nth level of H atom. If all the H-atoms are in the ground state and radiation of frequency $\frac{{\mathit{E}}_2-{\mathit{E}}_1}{\mathit{h}}$  falls on it

(a) It will not be absorbed at all

(b) Some of atoms will move to first excited state

(c) All atoms will be excited to the n= 2 state

(d) No atoms will make a transition to then n=3 state

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Answer-(b,d)

Explanation- when a radiation of energy fall on it some atoms would be excited but not all would be excite also no atom will go to 3 level some go to 2 level also but not all excite to 2 level. So by considering these facts we can say b and d option are correct

13. The simple Bohr model is not applicable to He⁴atom because
(a) He⁴ is an inert gas 
(b) He⁴ has neutrons in the nucleus
(c) He⁴ has one more electron
(d) Electrons are not subject to central forces

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Answer-(c,d)

Explanation- The simple Bohr model is not applicable to He⁴ atom because He⁴ has one more electron and electrons are not subject to central forces.