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

According to Einstein's theory of photoelectric effect, a single incident photon ejects a single electron. Therefore, when intensity increases, the number of incident photons increases, so number of ejected electrons increases, hence, photocurrent increases.
Now, maximum energy of electron $$ = \frac{1}{2}mv_{\max }^2$$   and $$\frac{1}{2}mv_{\max }^2 = e{V_0},$$    where $${V_0}$$ is stopping potential.
Thus, the maximum kinetic energy of the electrons does not depend upon the intensity of the incident rays, because the stopping potential is not affected by the increase of the intensity of rays. Hence, options (C) and (D) are wrong.

$$\eqalign{ & E = {W_0} + {K_{\max }}\,......\left( {\text{i}} \right) \cr & \Rightarrow hf = {W_A} + {K_A}\,.......\left( {{\text{ii}}} \right) \cr & {\text{and}}\,\,2hf = {W_B} + {K_B} = 2{W_A} + {K_B} \cr & \left( {\because \frac{{{W_A}}}{{{W_B}}} = \frac{1}{2}} \right) \cr} $$
Dividing equation (i) by (ii)
$$\frac{1}{2} = \frac{{{W_A} + {K_A}}}{{2{W_A} + {K_B}}} \Rightarrow \frac{{{K_A}}}{{{K_B}}} = \frac{1}{2}$$

Kinetic energy in photoelectric effect can also be written as
$$\eqalign{ & KE = \phi - {\phi _0}\,\,\left[ {_{{\phi _0} = \,{\text{work function}}}^{\phi \, = \,{\text{incident energy}}}} \right] \cr & {\text{Given,}}\,\,K{E_1} = 1 - 0.5 = 0.5\,eV \cr & K{E_2} = 2.5 - 0.5 = 2\,eV \cr & {\text{So,}}\,\,\frac{{K{E_1}}}{{K{E_2}}} = \frac{{0.5}}{2} = \frac{1}{4}\,\,{\text{or}}\,\,\frac{{v_1^2}}{{v_2^2}} = \frac{1}{4}\,\,\left[ {\because KE = \frac{1}{2}m{v^2}} \right] \cr & {\text{or}}\,\,\frac{{{v_1}}}{{{v_2}}} = \sqrt {\frac{1}{4}} = \frac{1}{2} \cr} $$

Energy of photon is given by
$$\eqalign{ & E = \frac{{hc}}{\lambda } = \frac{{12375}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}}\,eV\,\,\left[ {\because hc = 12375\,eV - \mathop {\text{A}}\limits^ \circ } \right] \cr & \therefore E = \frac{{12375}}{{5000}} \cr & = 2.48\,eV \cr} $$
According to Einstein’s photoelectric equation
$$\eqalign{ & KE = E - {W_0} = 2.48\,eV - 1.9\,eV \cr & = 0.58\,eV \cr} $$

The continuous X-ray spectrum is shown in figure.
All wavelengths $$ > {\lambda _{\min }}$$   are found, where $${\lambda _{\min }} = \frac{{12400}}{{V\left( {{\text{in}}\,{\text{volts}}} \right)}}\mathop {\text{A}}\limits^ \circ $$
Here $$V$$ is the applied voltage.

It is found that the photoelectric current increases linearly with the intensity of incident light.

It means number of photoelectrons emitted per second from photosensitive plate is directly proportional to the intensity of the incident radiation.

According to Einstein's photoelectric effect,
$$\eqalign{ & eV = \frac{{hc}}{\lambda } - \frac{{hc}}{{{\lambda _0}}}\,......\left( {\text{i}} \right) \cr & \frac{{eV}}{4} = \frac{{hc}}{{2\lambda }} - \frac{{hc}}{{{\lambda _0}}}\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing equation (i) by (ii) by
$$ \Rightarrow 4 = \frac{{\frac{1}{\lambda } - \frac{1}{{{\lambda _0}}}}}{{\frac{1}{{2\lambda }} - \frac{1}{{{\lambda _0}}}}}$$
on solving we get,
$${\lambda _0} = 3\lambda $$

By using $$hv - h{v_0} = {K_{\max }}$$
$$\eqalign{ & \Rightarrow h\left( {{v_1} - {v_0}} \right) = {K_1}\,.......\left( {\text{i}} \right) \cr & {\text{And}}\,h\left( {{v_2} - {v_0}} \right) = {K_2}\,......\left( {{\text{ii}}} \right) \cr & \Rightarrow \frac{{{v_1} - {v_0}}}{{{v_2} - {v_0}}} = \frac{{{K_1}}}{{{K_2}}} = \frac{1}{K}, \cr} $$
Hence $${v_0} = \frac{{k{v_1} - {v_2}}}{{K - 1}}.$$

As a photon moves with the velocity of light
i.e. $$\nu = c$$
∴ The momentum of the photon $$ = mc = \frac{{h\nu }}{c} = \frac{h}{\lambda }\,\,\left[ {\lambda = \frac{c}{\nu }} \right]$$

From Einstein's photoelectric equation
$$K.E{._\lambda } = \frac{{hc}}{\lambda } - \phi \,.......\left( {\text{i}} \right)\,\,\left( {{\text{for monochromatic light of wavelength }}\lambda } \right)$$
where $$\phi $$ is work function
$$K.E{._{\frac{\lambda }{2}}} = \frac{{hc}}{{\frac{\lambda }{2}}} - \phi \,.......\left( {{\text{ii}}} \right)\,\,\left( {{\text{for monochromatic light of wavelength }}\frac{\lambda }{2}} \right)$$
From question,
$$\eqalign{ & K.E{._{\frac{\lambda }{2}}} = 3\left( {K.E{._\lambda }} \right) \cr & \Rightarrow \frac{{hc}}{{\frac{\lambda }{2}}} - \phi = 3\left( {\frac{{hc}}{\lambda } - \phi } \right) \cr & \frac{{2hc}}{\lambda } - \phi = 3\frac{{hc}}{\lambda } - 3\phi \cr & \Rightarrow 2\phi = \frac{{hc}}{\lambda } \cr & \therefore \phi = \frac{{hc}}{{2\lambda }} \cr} $$