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SolveItClass 12 · Exemplar

NCERT Exemplar · Class 12 Physics Atoms

29 questions

Very Short Answer Questions 14–18 (part 3 of 5)

  1. Exercise 14

    The mass of a H-atom is less than the sum of the masses of a proton and electron. Why is this?
    NCERT’s answer
    Einstein's mass-energy equivalence gives \(\displaystyle E=m c^2\). Thus the mass of a H-atom is \(\displaystyle m_p+m_e-\frac{B}{c^2}\) where \(\displaystyle \mathrm{B} \approx 13.6 \mathrm{eV}\) is the binding energy.
    \[m_H = m_p + m_e - \frac{B}{c^2} \] \[B = 13.6\ \mathrm{eV} = 13.6\times1.6\times10^{-19}\ \mathrm{J} = 2.18\times10^{-18}\ \mathrm{J} \] \[\Delta m = \frac{B}{c^2} = \frac{2.18\times10^{-18}\ \mathrm{J}}{(3\times10^{8}\ \mathrm{m\,s^{-1}})^2} \approx 2.4\times10^{-35}\ \mathrm{kg} \] Answer: The atom gave out its binding energy ($\displaystyle 13.6$ eV) on forming, so its mass is less by the mass defect \(\displaystyle B/c^2\).

  2. Exercise 15

    Imagine removing one electron from He4\displaystyle \mathrm{He}^4 and He3\displaystyle \mathrm{He}^3. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.
    NCERT’s answer
    Because both the nuclei are very heavy as compared to electron mass.
    \[E_n = -\frac{\mu Z^2 e^4}{8\varepsilon_0^2 h^2}\,\frac{1}{n^2},\qquad \mu = \frac{m_e M}{m_e + M} = \frac{m_e}{1 + m_e/M} \] \[Z = 2 \ \text{for both } \mathrm{He}^{+} \text{ ions} \] \[\frac{m_e}{M_4} \approx \frac{1}{4\times1836} = 1.4\times10^{-4},\qquad \frac{m_e}{M_3} \approx \frac{1}{3\times1836} = 1.8\times10^{-4} \Rightarrow \mu \approx m_e \] \[\frac{\mu_4}{\mu_3} \approx 1 + 4.5\times10^{-5},\qquad E_n \approx -\frac{54.4\ \mathrm{eV}}{n^2} \ \text{for both} \] Answer: Both ions have \(\displaystyle Z = 2\) and a nucleus far heavier than the electron, so the reduced mass is almost \(\displaystyle m_e\) in each.

  3. Exercise 16

    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?
    NCERT’s answer
    Because electrons interact only electromagnetically.
    \[h\nu = E_i - E_f \] Answer: The electron interacts only electromagnetically, so the energy can leave only as electromagnetic radiation, a photon.

  4. Exercise 17

    Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+4/3)e\displaystyle (+4 / 3) e and electron a charge (−3/4)e\displaystyle (-3 / 4) e, where e=1.6×10−19C\displaystyle e=1.6 \times 10^{-19} \mathrm{C}. Give reasons for your answer.
    NCERT’s answer
    Yes, since the Bohr formula involves only the product of the charges.
    \[F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_p\,q_e}{r^2},\qquad q_p q_e = \left(\frac{4}{3}e\right)\left(\frac{3}{4}e\right) = e^2 \] \[\frac{m v^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2},\qquad m v r = \frac{nh}{2\pi} \] \[E_n = -\frac{m e^4}{8\varepsilon_0^2 h^2 n^2} \quad (\text{same as before}) \] Answer: Yes, the formula is unchanged, because only the product of the two charges enters and it is still \(\displaystyle e^2\).

  5. Exercise 18

    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?
    NCERT’s answer
    No, because accoding to Bohr model, \(\displaystyle E_n=-\frac{13.6}{n^2}\), and electons having different energies belong to different levels having different values of \(\displaystyle n\). So, their angular momenta will be different, as \(\displaystyle m v r=\frac{n h}{2 \pi}\).
    \[E_n = -\frac{13.6}{n^2}\ \mathrm{eV},\qquad L = \frac{nh}{2\pi} \] Atoms \(\displaystyle A\) and \(\displaystyle B\): \[E_A \neq E_B \Rightarrow n_A \neq n_B \Rightarrow L_A \neq L_B \] Answer: No. Different energies mean different \(\displaystyle n\), hence different angular momenta.