Question

A photon of wavelength $5000\mathring{A}$ strikes a metal surface having work function of $2.20eV$. The kinetic energy of the emitted photoelectron is

• A. $4.4\times 10^{-20}J$
• B. $0.425eV$
• C. $2.20eV$
• D. no electron will be emitted

Answer 1

Answer: (A)

Energy of photon, $E=hv=\frac{hc}{\lambda }$

$=\frac{\left(6.6\times 10^{-34}Js\right)\left(3\times 10^{8}m{s}^{-1}\right)}{5000\times 10^{-10}m}$

$=3.96\times 10^{-19}J$

$=\frac{3.96\times 10^{-19}J}{1.6\times 10^{-19}J/eV}=2.475eV$

$\left(\because 1ev=1.6\times 10^{-19}J\right)$

Kinetic energy of the emitted photon

$=hv-h{v}_0$ $=2.475-2.20=0.275eV$

$=0.275\times 1.6\times 10^{-19}J=4.4\times 10^{-20}J$