Question

Photoeelectric work-function of a metal is $1eV.$ Light of wavelength $\lambda =3000\mathring{A}$ falls on it. The photoelectrons come out with maximum velocity:

• A. $10m/s$
• B. ${10}^{3}m/s$
• C. ${10}^{4}m/s$
• D. ${10}^{6}m/s$

Answer 1

The correct option is D ${10}^{6}m/s$

Solution

By Binstein's photo electric $e{q}^n.$

$hr=h{r}_0+K{E}_{max}$

$\Rightarrow \frac{hc}{h}=h{r}^0+\frac{1}{2}m{V}^2$

Here,

$\Rightarrow \frac{6.6\times {10}^{-31}\times 3\times {10}^8}{3\times {10}^{-7}}=1.0\times {10}^{-19}J+\frac{9.k{10}^{-31}}{z}\times V^2$

$\Rightarrow 6.6\times {10}^{-19}-1.6\times {10}^{-19}=\frac{9.1\times {10}^{-31}}{2}{V}^2$

$\Rightarrow 2\times \frac{5\times {10}^{-19}}{9.1\times {10}^{-31}}=V^2$

$\Rightarrow V=\sqrt{\frac{{10}^{13}}{9.1}}$

$\Rightarrow V\approx {10}^{6}m/s$

option -D is correct.