Question

The work function of a metal surface is 1 eV. A light of wavelength 3000$A^0$ is incident on it. The maximum velocity of the photo-electrons is nearly

• A. ${10}^{6}m{s}^{-1}$
• B. ${10}^{4}m{s}^{-1}$
• C. ${10}^{2}m{s}^{-1}$
• D. $10m{s}^{-1}$

Answer 1

The correct option is A ${10}^{6}m{s}^{-1}$

The maximum K.E with which electrons comes out,

$\frac{hc}{\lambda }-1eV=\frac{1}{2}m{v}^2$

$\Rightarrow \frac{4.14}{{10}^{-10}}-1eV=\frac{1}{2}m\times v^2$

$\Rightarrow \frac{2\times 3.14\times 1.6\times {10}^{-19}\times {10}^{31}}{9.1}=V^2$

$V^2=1.104175\times {10}^{12}$

$\Rightarrow V^2=1.005\times {10}^6$

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

So, the answer is option (A).