Q. The work function of metal is $1\ eV.$ Light of wavelength $3000\ \mathring{\mathrm{A}}$ is incident on this metal surface. The velocity of emitted photo-electrons will be
A) $10\ m/s$
B) $1 \times 10^{3}\ m/s$
C) $1 \times 10^{4}\ m/s$
D) $1 \times 10^{6}\ m/s$
Explanation
(d) $E = W_{0} + K_{\max};E = \frac{12375}{3000} = 4.125\ eV$
$\Rightarrow K_{\max} = E - W_{0} = 4.125\ eV - 1\ eV = 3.125\ eV$
$\Rightarrow \frac{1}{2}mv_{\max}^{2} = 3.125 \times 1.6 \times 10^{- 19}J$
$\Rightarrow v_{\max} = \sqrt{\frac{2 \times 3.125 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}} = 1 \times 10^{6}m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.