Question

The work function of a metal is $1eV$ . Light of wavelength $3000\mathring{A}$ is incident on this metal surface. The velocity of emitted photoelectrons will be

• A. $10m{s}^{-1}$
• B. $10^{3}m{s}^{-1}$
• C. $10^{4}m{s}^{-1}$
• D. $10^{6}m{s}^{-1}$

Answer 1

Answer: (D)

$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-{\varphi }_0$
$KE$ in $eV$ is $K=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\left(3\times 10^{-7}\right)\times 1.6\times 10^{-19}}-1$
$=4.14-1=3.14eV$
or $v=\sqrt{\frac{2\times 3.14\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}}=10^{6}m{s}^{-1}$