Atoms And Nuclei MCQ Questions & Answers in Modern Physics | Physics

The circumference of an orbit in an atom in terms of wavelength of wave associated with electron is given by
$$2\pi {r_n} = n\lambda \,\,\left[ {{r_n} = {\text{radius}}\,{\text{of}}\,{\text{any}}\,n\,{\text{orbit}}} \right]$$

The energy of the system of two atoms of diatomic molecule $$E = \frac{1}{2}I{\omega ^2}$$
where $$I$$ = moment of inertia
$$\omega = {\text{Angular}}\,{\text{velocity}} = \frac{L}{I},$$
$$L$$ = Angular momentum
$$I = \frac{1}{2}\left( {{m_1}r_1^2 + {m^2}r_2^2} \right)$$
Thus, $$E = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right){\omega ^2}\,......\left( {\text{i}} \right)$$
$$\eqalign{ & E = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)\frac{{{L^2}}}{{{I^2}}} \cr & L = n\frac{{nh}}{{2n}}\,\left( {{\text{According}}\,{\text{Bohr's}}\,{\text{Hypothesis}}} \right) \cr & E = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)\frac{{{L^2}}}{{{{\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)}^2}}} \cr & E = \frac{1}{2}\frac{{{L^2}}}{{\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)}} = \frac{{{n^2}{h^2}}}{{8{\pi ^2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)}} \cr & E = \frac{{\left( {{m_1} + {m_2}} \right){n^2}{h^2}}}{{8{\pi ^2}{r^2}{m_1}{m_2}}}\left[ {\because {r_1} = \frac{{{m_2}r}}{{{m_1} + {m_2}}};{r_2} = \frac{{{m_2}r}}{{{m_1} + {m_2}}}} \right] \cr} $$

KEY CONCEPT:
$$\lambda \propto \frac{1}{m}$$
For ordinary hydrogen atom, longest wavelength
$$\frac{1}{\lambda } = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right] = \frac{{5R}}{{36}}\,{\text{or}}\,\lambda = \frac{{36}}{{5R}}$$
With hypothetical particle, required wavelength
$$\lambda ' = \frac{1}{2} \times \frac{{36}}{{5R}} = \frac{{18}}{{5R}}$$

We have to find the frequency of emitted photons. For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options (c) and (d).
Frequency is given by
$$h\nu = - 13.6\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)$$
For transition from $$n = 6$$  to $$n = 2,$$
$${\nu _1} = \frac{{ - 13.6}}{h}\left( {\frac{1}{{{6^2}}} - \frac{1}{{{2^2}}}} \right) = \frac{2}{9} \times \left( {\frac{{13.6}}{h}} \right)$$
For transition from $$n = 2$$  to $$n = 1,$$
$$\eqalign{ & {\nu _2} = \frac{{ - 13.6}}{h}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{1^2}}}} \right) = \frac{3}{4} \times \left( {\frac{{13.6}}{h}} \right) \cr & \therefore {\nu _1} < {\nu _2} \cr} $$

Radius of circular path followed by electron is given by,
$$\eqalign{ & r = \frac{{m\upsilon }}{{qB}} = \frac{{\sqrt {2meV} }}{{eB}} = \frac{1}{B}\sqrt {\frac{{2m}}{e}V} \cr & \Rightarrow V = \frac{{{B^2}{r^2}e}}{{2m}} = 0.8V \cr} $$
For transition between 3 to 2.
$$E = 13.6\left( {\frac{1}{4} - \frac{1}{9}} \right) = \frac{{13.6 \times 5}}{{36}} = 1.88eV$$
Work function $$ = 1.88eV - 0.8eV$$
$$ = 1.08eV \approx 1.1eV$$

When $$F = \frac{k}{r} = $$   centripetal force, then $$\frac{k}{r} = \frac{{m{v^2}}}{r}$$
$$ \Rightarrow m{v^2} = {\text{constant}} \Rightarrow $$     kinetic energy is constant
$$ \Rightarrow T$$  independent of $$n.$$

Wavelength for Lyman series
$${\lambda _L} = \frac{1}{{R\left( {1 - \frac{1}{4}} \right)}} = \frac{4}{{3R}}$$
and wavelength for Balmer series
$$\eqalign{ & {\lambda _B} = \frac{1}{{R\left( {\frac{1}{4} - \frac{1}{9}} \right)}} = \frac{1}{{R\left( {\frac{5}{{36}}} \right)}} = \frac{{36}}{{5R}} \cr & \therefore \frac{{{\lambda _L}}}{{{\lambda _B}}} = \frac{4}{{3R}} \times \frac{{5R}}{{36}} = \frac{5}{{27}} \Rightarrow {\lambda _L}:{\lambda _B} = 5:27 \cr} $$

According to Bohr’s hypothesis, electron can revolve only in those orbits in which its angular momentum is an integral multiple of $$\frac{h}{{2\pi }},$$  where $$h$$ is Planck’s constant. In these orbits, angular momentum of electron can have magnitude as
$$\frac{h}{{2\pi }},\frac{{2h}}{{2\pi }},\frac{{3h}}{{2\pi }}......$$    etc.but never as $$\frac{{1.5\,h}}{{2\pi }},\frac{{2.5\,h}}{{2\pi }},\frac{{3.5\,h}}{{2\pi }}......$$     etc. This is called the quantisation of angular momentum.

From energy level diagram, using $$\Delta E = \frac{{hc}}{\lambda }$$
For wavelength $${\lambda _1}\Delta E = - E - \left( { - 2E} \right) = \frac{{hc}}{{{\lambda _1}}}$$
$$\therefore {\lambda _1} = \frac{{hc}}{E}$$
For wavelength $${\lambda _2}\Delta E = - E - \left( { - \frac{{4E}}{3}} \right) = \frac{{hc}}{{{\lambda _2}}}$$
$$\therefore {\lambda _2} = \frac{{hc}}{{\left( {\frac{E}{3}} \right)}}\,\,\,\,\,\,\,\therefore r = \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{1}{3}$$

Given, $${E_n} = - \frac{{13.6}}{{{n^2}}}eV$$
Energy of photon ejected when electron jumps from $$n = 3$$  to $$n = 2$$  state is given by
$$\Delta E = {E_3} - {E_2}$$
Energy of third orbit
$${E_3} = - \frac{{13.6}}{{{{\left( 3 \right)}^2}}}eV = - \frac{{13.6}}{9}eV$$
Energy of second orbit
$${E_2} = - \frac{{13.6}}{{{{\left( 2 \right)}^2}}}eV = - \frac{{13.6}}{4}eV$$
$${\text{So,}}\,\,\Delta E = {E_3} - {E_2} = - \frac{{13.6}}{9} - \left( { - \frac{{13.6}}{4}} \right)$$
$$ = 1.9\,eV$$  (approximately)