Dual Nature of Matter and Radiation MCQ Questions & Answers in Modern Physics | Physics

The cut off wavelength is given by
$${\lambda _0} = \frac{{hc}}{{eV}}\,.......\left( {\text{i}} \right)$$
According to de Broglie equation
$$\eqalign{ & \lambda = \frac{h}{p} = \frac{h}{{\sqrt {2meV} }} \cr & \Rightarrow {\lambda ^2} = \frac{{{h^2}}}{{2meV}} \cr & \Rightarrow V = \frac{{{h^2}}}{{2me{\lambda ^2}}}\,.......\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii),
$${\lambda _0} = \frac{{hc \times 2me{\lambda ^2}}}{{e{h^2}}} = \frac{{2mc{\lambda ^2}}}{h}$$

When an electron in hydrogen atom jumps from first excited state $$\left( {n = 2} \right)$$  to ground state $$\left( {n = 1} \right)$$  energy is released and is given by
$$E = {E_{\left( {n = 2} \right)}} - {E_{\left( {n = 1} \right)}}$$
where, $${E_n} = - \frac{{13.6}}{{{n^2}}}eV$$
Energy released from emission of electron is given by
$$\eqalign{ & E = - 3.4 - \left( { - 13.6} \right) \cr & = 10.2\,eV \cr} $$
Now, from photoelectric equation, work function,
$$\eqalign{ & \phi = E - eV = h\nu \cr & \nu = \frac{{E - eV}}{h} = \frac{{\left( {10.2 - 3.57} \right)e}}{{6.67 \times {{10}^{ - 34}}}} \cr & \nu = \frac{{6.63 \times 1.6 \times {{10}^{ - 19}}}}{{6.67 \times {{10}^{ - 34}}}} \cr & = 1.6 \times {10^{15}}\,Hz \cr} $$

Applying conservation of linear momentum,
Initial momentum = Final momentum
$$\eqalign{ & 0 = {m_1}{v_1} - {m_2}{v_2} \cr & \Rightarrow {m_1}{v_1} = {m_2}{v_2} \cr & {\text{Now,}}\frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{\frac{h}{{{m_1}{v_1}}}}}{{\frac{h}{{{m_2}{v_2}}}}} = 1 \cr} $$

When electrons are accelerated through $$V$$ volt, the gain in $$KE$$  of the electron is given by
$$\eqalign{ & \frac{1}{2}m{v^2} = eV\,\,{\text{or}}\,\,{v^2} = \frac{{2eV}}{m} \cr & {\text{Given,}}\,\,V = 100\;V \cr & {\text{So,}}\,\,v = \sqrt {\frac{{2e\left( {100} \right)}}{m}} \,......\left( {\text{i}} \right) \cr} $$
According to de-Broglie theory, wavelength
$$\eqalign{ & \lambda = \frac{h}{{mv}} \cr & \therefore \lambda = \frac{h}{{m\sqrt {\frac{{2\,e\left( {100} \right)}}{m}} }} \cr & = \frac{h}{{\sqrt {2me\left( {100} \right)} }} \cr & = \frac{{6.6 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 100} }} \cr & = 1.2 \times {10^{ - 10}} \cr & \Rightarrow 1.2\,\mathop {\text{A}}\limits^ \circ \cr} $$

From Bragg’s equation
$$\eqalign{ & d\sin \theta = \lambda \cr & \sin \theta = \frac{\lambda }{d} < 1 \cr & \therefore \lambda < d \cr & \frac{h}{{\left| {{p_y}} \right|}} < d\,\,\left[ {\because \lambda = \frac{h}{{\left| {{p_y}} \right|}}} \right] \cr & \therefore h < \left| {{p_y}} \right|d \cr} $$

The energy of incident photons is given by
$$hv = e{V_s} + {\phi _0} = 2 + 5 = 7\,eV\,\,\left( {{V_s}\,{\text{is stopping potential and }}{\phi _0}\,{\text{is}}\,{\text{work}}\,{\text{function}}} \right)$$
Saturation current $$ = {10^{ - 5}}A$$
$$\eqalign{ & = \frac{{\eta P}}{{hv}}e = \frac{{{{10}^{ - 5}}P}}{{7 \times e}}e\left( {\eta \,{\text{is photon emission efficiency}}} \right) \cr & \therefore P = 7W \cr} $$

According to the concept of threshold minimum frequency needed for photoelectric emission i.e.
$$\eqalign{ & KE = h\nu - {\phi _0} \cr & {\text{or}}\,\,\frac{1}{2}m{\nu ^2} = h\nu - h{\nu _0} \cr & {\text{So,}}\,\nu \geqslant {\nu _0} \cr} $$

For one photocathode
$$h{f_1} - W = \frac{1}{2}mv_1^2\,.......\left( {\text{i}} \right)$$
For another photo cathode
$$h{f_2} - W = \frac{1}{2}mv_2^2\,.......\left( {{\text{ii}}} \right)$$
Subtracting (ii) from (i) we get
$$\eqalign{ & \left( {h{f_1} - W} \right) - \left( {h{f_2} - W} \right) = \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 \cr & \therefore h\left( {{f_1} - {f_2}} \right) = \frac{m}{2}\left( {v_1^2 - v_2^2} \right) \cr & \therefore v_1^2 - v_2^2 = \frac{{2h}}{m}\left( {{f_1} - {f_2}} \right) \cr} $$

In J J Thomson's method, as the electron beam is accelerated from cathode to anode, its potential energy at the cathode appears as gain in kinetic energy at the anode. If $$V$$ is the potential difference between cathode and anode, then potential energy of electron at cathode $$ = {\text{charge}} \times {\text{potential difference}} = eV$$
Gain in kinetic energy of electron at anode $$ = \frac{1}{2}m{v^2}$$
According to conservation of energy, we have
$$\eqalign{ & eV = \frac{1}{2}m{v^2} \cr & \therefore v = \sqrt {\left( {\frac{{2eV}}{m}} \right)} \cr} $$

$$\eqalign{ & E = \frac{{hc}}{\lambda } = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1 \times {{10}^{ - 10}}}} \cr & = 2 \times {10^{ - 15}}J \cr} $$