Modern Physics Miscellaneous MCQ Questions & Answers in Modern Physics | Physics

We know that
$$e{V_0} = {K_{\max }} = h\nu - \phi $$
where, $$\phi $$ is the work function.
Hence, as $$\nu $$ increases (note that frequency of X-rays is greater than that of U.V. rays), both $${V_0}$$ and $${K_{\max }}$$  increase. So statement - 1 is correct

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
$$\eqalign{ & \left( {h{f_1} - W} \right) - \left( {h{f_2} - W} \right) = \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 \cr & \therefore h\left( {{f_1} - {f_2}} \right) = \frac{m}{2}\left( {v_1^2 - v_2^2} \right) \cr & \therefore v_1^2 - v_2^2 = \frac{{2h}}{m}\left( {{f_1} - {f_2}} \right) \cr} $$

$$\frac{{{\lambda _{Cu}}}}{{{\lambda _{Mo}}}} = \frac{{{{\left( {{Z_{Mo}} - 1} \right)}^2}}}{{{{\left( {{Z_{Cu}} - 1} \right)}^2}}} = {\left( {\frac{{42 - 1}}{{29 - 1}}} \right)^2} = {\left( {\frac{{41}}{{28}}} \right)^2} = 2.14$$
\[\left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\therefore \sqrt \nu = \left( {Z - b} \right)\,{\text{here }}b = 1} \\ {\nu = {{\left( {Z - 1} \right)}^2}} \end{array}} \\ {\frac{1}{\lambda } \propto {{\left( {Z - 1} \right)}^2}} \end{array}} \right]\]

Note:
Stopping potential is the negative potential applied to stop the electrons having maximum kinetic energy. Therefore, stopping potential will be $$4\,volt.$$

In X-ray tube, $${\lambda _{\min }} = \frac{{hc}}{{eV}}$$
In $${\lambda _{\min }} = In\left( {\frac{{hc}}{e}} \right) - InV$$
Clearly, $$\log {\lambda _{\min }}$$  versus log $$V$$ graph
slope is negative hence option (C) correctly depicts.

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

$$\eqalign{ & I \propto \frac{1}{{{r^2}}};\frac{{{I_1}}}{{{I_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} = \frac{1}{4} \cr & {I_2} \to 4\,{\text{times}}\,{I_1} \cr} $$
When intensity becomes 4 times, no. of photoelectrons emitted would increase by 4 times, since number of electrons emitted per second is directly proportional to intensity.

We know that work function is the energy required and energy $$E = h\upsilon $$
$$\eqalign{ & \therefore \frac{{{E_{Na}}}}{{{E_{Cu}}}} = \frac{{h{\upsilon _{Na}}}}{{h{\upsilon _{Cu}}}} = \frac{{{\lambda _{Cu}}}}{{{\lambda _{Na}}}}\,\,\,\,\,\left[ {\because \upsilon \propto \frac{1}{\lambda }\,{\text{for}}\,{\text{light}}} \right] \cr & \therefore \frac{{{\lambda _{Na}}}}{{{\lambda _{Cu}}}} = \frac{{{E_{Cu}}}}{{{E_{Na}}}} = \frac{{4.5}}{{2.3}} \approx \frac{2}{1} \cr} $$

The ionisation potential increases from left to right in a period and decreases from top to bottom in a group. Therefore ceasium will have the lowest ionisation potential.

KEY CONCEPT :
$$\eqalign{ & {\lambda _{\min }} = \frac{{hc}}{E} \cr & \therefore {\lambda _{\min }} = \frac{{12400}}{{80 \times {{10}^3}}}\mathop {\text{A}}\limits^ \circ = 0.155\mathop {\text{A}}\limits^ \circ \cr} $$
Energy of incident electrons is greater than the ionization energy of electrons in $$K$$-shell, the $$K$$-shell electrons will be knocked off. Hence, characteristic X-ray spectrum will be obtained.