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

Einstein's photoelectric equation is
$$KE = hv - {W_0}\,......\left( {\text{i}} \right)$$
where, $${W_0}$$ = work function of metal, $$\left( { = h{v_0}} \right)$$
Comparing above Eq. (i) with equation of a straight line $$y = mx + c$$
we get, $$m = h,c = - {W_0}$$
Therefore, if we draw a graph between kinetic energy and frequency, then a straight line cutting the frequency axis at $${{v_0}}$$ and giving an intercept of $$\left( { - {W_0}} \right)$$  on the kinetic energy axis, is obtained.
NOTE
As we know that for emission of electrons, there is a certain threshold frequency after which emission starts. Considering this fact graph (D) is correct.

We know that,
de-Broglie wavelength $$\lambda = \frac{h}{P} = \frac{h}{{\sqrt {2m\left( {KE} \right)} }}$$
$$K.E.$$  of thermal neutron $$ = \frac{3}{2}kT$$
$$ = \frac{h}{{\sqrt {2m\left( {\frac{3}{2}kT} \right)} }} \Rightarrow \lambda = \frac{h}{{\sqrt {3mkT} }}$$

Efficient power $$\left( P \right)$$ is given by
$$\eqalign{ & P = \frac{N}{t} \times \frac{{hc}}{\lambda }\,\,\left( {N = {\text{total number of photons}}} \right) \cr & \frac{N}{t} = \frac{{P \times \lambda }}{{hc}} \cr & = \frac{{50 \times 0.6 \times {{10}^{ - 6}}}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}\,\,\left[ {\because \, = 25\% \,{\text{of}}\,200W = 50W} \right] \cr & = 1.5 \times {10^{20}} \cr} $$

$$\eqalign{ & {\text{Here,}}\,\lambda = 667 \times {10^{ - 9}}m,P = 9 \times {10^{ - 3}}W \cr & {\text{Power}} = \frac{{{\text{energy}}\left( E \right)}}{{{\text{time}}\left( t \right)}} \cr & = \frac{{nhc}}{{\lambda t}} = \frac{{Nhc}}{\lambda }\,\,\left[ {_{N = \,{\text{Number of photons emitted per second}}\, = \,\,\frac{n}{t}}^{E = \,\,\frac{{hc}}{\lambda },\,n\,\, = \,{\text{Total number of photons}}}} \right] \cr & {\text{So,}}\,\,N = \frac{{P \times \lambda }}{{hc}} = \frac{{9 \times {{10}^{ - 3}} \times 667 \times {{10}^{ - 9}}}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \cr & = 3 \times {10^{16}}\,m/s \cr} $$

According to photo-electric equation :
$$K.{E_{\max }} = hv - h{v_0}\left( {{\text{Work function}}} \right)$$
Some sort of energy is used in ejecting the photoelectrons.

For photo electric equation,
Einstein's equation can be written as
$${\left( {KE} \right)_{\max }} = h\nu - {\phi _0}$$
For the first condition,
$$0.5 = E - {\phi _0}\,........\left( {\text{i}} \right)$$
For the second condition,
$$0.8 = 1.2E - {\phi _0}\,........\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii)
$$\eqalign{ & - 0.3 = - 0.2E \cr & E = \frac{{0.3}}{{0.2}} = 1.5\,eV \cr} $$
From Eq. (i),
$$\eqalign{ & 0.5 = 1.5 - {\phi _0} \cr & {\phi _0} = 1.5 - 0.5 \cr & = 1\,eV \cr} $$

Saturation current is inversely proportional to the square of distance of cathode from point source.

From photo-electric equation,
$$\eqalign{ & h\nu = h\nu + {K_{\max }}\,.......\left( {\text{i}} \right) \cr & h.2\nu = h\nu + \frac{1}{2}mV_{\max }^2\,\,\left[ {\therefore \nu ' = 2\nu } \right] \cr & \Rightarrow h\nu = \frac{1}{2}mV_{\max }^2 \cr & \Rightarrow {V_{\max }} = \sqrt {\frac{{2h\nu }}{m}} \cr} $$

Power emitted,
$$P = 2 \times {10^{ - 3}}W$$
Energy of photon,
$$E = h\nu = 6.6 \times {10^{ - 34}} \times 6 \times {10^{14}}J$$
Here, $$h$$ being Planck's constant.
Number of photons emitted per second is given by
$$\eqalign{ & n = \frac{{{\text{Power}}\left( P \right)}}{{{\text{Energy}}\left( E \right)}} \cr & = \frac{{2 \times {{10}^{ - 3}}}}{{6.6 \times {{10}^{ - 34}} \times 6 \times {{10}^{14}}}} \cr & = 5 \times {10^{15}} \cr} $$

For electron De-Broglie wavelength,
$${\lambda _e} = \frac{h}{{\sqrt {2mE} }};{\text{For}}\,{\text{photon}}\,E = pc$$
⇒ De-Broglie wavelength, $${\lambda _{{\text{Ph}}}} = \frac{{hc}}{E}$$
$$\therefore \frac{{{\lambda _e}}}{{{\lambda _{{\text{Ph}}}}}} = \frac{h}{{\sqrt {2mE} }} \times \frac{E}{{hc}} = {\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}\frac{1}{c}$$