Physics 3 Marks Question

We get, $\frac{\text{hc}}{\lambda}+\text{m}_0\text{c}^2=\text{mc}^2$

and applying conservation of momentum $\frac{\text{h}}{\lambda}=\text{mv}$

Mass of e $=\text{m}=\frac{\text{m}_0}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}$

from above equation it can be easily shown that

V = C

V = 0

both of these results have no physical meaning hence it is not possible for a photon to be completely absorbed by a free electron.

$\text{Energy}=\frac{\text{kq}^2}{\text{R}}=\frac{\text{kq}^2}{1}$

Now, $\frac{\text{kq}^2}{1}=\frac{\text{hc}}{\lambda}$

$\lambda=\frac{\text{hc}}{\text{kq}^2}$

For max $'\lambda',$ ‘q’ should be min,

For minimum ‘e’ = 1.6 × 10^-19C

$\text{Max}\lambda=\frac{\text{hc}}{\text{kq}^2}=0.863\times10^3=863\text{m.}$

For next smaller wavelength

$=\frac{6.63\times3\times10^{-34}\times10^{8}}{9\times10^9\times(1.6\times2)^2\times10^{-38}}$

$\frac{863}{4}=215.74\text{m}$

$\therefore$ Frequency $=\frac{1.2\times10^{15}}{2}=0.6\times10^{15}$

$\text{hv}=\phi_0+\text{kE}$

$\Rightarrow\text{hv}-\phi_0=\text{KE}$

$\Rightarrow\text{KE}=\frac{6.63\times10^{-34}\times0.6\times10^{15}}{1.6\times10^{-19}}-2$

$=0.482\text{ev}=0.48\text{ev.}$

$\text{E}=\text{mc}^2$

^{$\Rightarrow\text{E}=\frac{\text{m}_0\text{c}^2}{\sqrt{1\frac{\text{v}^2}{\text{c}^2}}}$}

Relativistic momentum,

$\text{p}=\text{mv}$

$\Rightarrow\text{p}=\frac{\text{m}_0\text{c}^2}{\sqrt{1\frac{\text{v}^2}{\text{c}^2}}}$

Combining the above equations, we get:

$\text{E}^2=\text{m}_0^2\text{c}^ 4+\text{p}^2\text{c}^2$

From the above equation, it is clear that for $\text{p}=\frac{\text{E}}{\text{c}}$ to be valid, the rest mass of the body should be zero. As electrons do not have zero rest mass, this equation is not valid for electrons.

$=\frac{5\times10^{-3}}{8\times10^{15}}-1.6\times10^{-19}\times2$

$=6.25\times10^{-19}-3.2\times10^{-19}=3.05\times10^{-19}\text{J}$

$=\frac{3.05\times10^{-19}}{1.6\times10^{-19}}=1.906\text{ev.}$

$\text{p}=\frac{\text{h}}{\lambda}$

$\text{E}=\frac{\text{hc}}{\lambda}=\text{pc}$

$\Rightarrow\frac{\text{E}}{\text{t}}=\frac{\text{p}}{\text{t}}\text{c}$

$\text{pc}=\sqrt{\text{E}^2-\text{m}^2\text{c}^4.}$

A photon has zero rest mass. Therefore, on putting m = 0 in the equation, we get $\text{p}=\frac{\text{E}}{\text{c}},$ which is the valid equation for a photon.

1. We know W₀ = hv₀

$\text{v}_0=\frac{\text{W}_0}{\text{h}}=\frac{2.5\times10^{-19}}{6.63\times10^{-34}}$

$=3.77\times10^{14}\text{Hz}=3.8\times10^{14}\text{Hz}$

2. eV₀ = hv - W₀

or, $\text{v}_0=\frac{\text{hv}-\text{W}_0}{\text{e}}$

$=\frac{6.63\times10^{-34}\times6\times10^{14}-2.5\times10^{-19}}{1.6\times10^{-19}}=0.91\text{v}$

Now, $\text{Force}=\frac{\text{power}}{\text{velocity}}=\frac{60}{3\times10^8}=2\times10^{-7}\text{N}.$

$\text{pressure}=\frac{\text{force}}{\text{area}}=\frac{2\times10^{-7}}{4\times3.14\times(0.2)^2}$

$\frac{1}{8\times3.14}\times10^{-5}$

$=0.039\times10^{-5}=3.9\times10^{-7}=4\times10^{-7}\text{N/m}^2.$