Question

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

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

Answer 1

Answer: (D)

$\frac{1}{2}mv^{2}=\frac{h c}{\lambda }-\phi_{0}$
$KE$ 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 $v=\sqrt{\frac{2 \times 3.14 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}}=10^{6}ms^{- 1}$