Question

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

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

Answer 1

Answer: (A)

Energy of photon, $E=h v=\frac{h c}{\lambda}$

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

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

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

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

Kinetic energy of the emitted photon

$=h v-h v_{0}$ $=2.475-2.20=0.275\, eV$

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