# Light of frequency

Question:

Light of frequency $7.21 \times 10^{14} \mathrm{~Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5} \mathrm{~m} / \mathrm{s}$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Solution:

Frequency of the incident photon, $v=488 \mathrm{~nm}=488 \times 10^{-9} \mathrm{~m}$

Maximum speed of the electrons, v = 6.0 × 105 m/s

Planck’s constant, h = 6.626 × 10−34 Js

Mass of an electron, m = 9.1 × 10−31 kg

For threshold frequency ν0, the relation for kinetic energy is written as:

$\frac{1}{2} m v^{2}=h\left(v-v_{0}\right)$

$v_{0}=v-\frac{m v^{2}}{2 h}$

$=7.21 \times 10^{14}-\frac{\left(9.1 \times 10^{-31}\right) \times\left(6 \times 10^{5}\right)^{2}}{2 \times\left(6.626 \times 10^{-34}\right)}$

$=7.21 \times 10^{14}-2.472 \times 10^{14}$

$=4.738 \times 10^{14} \mathrm{~Hz}$

Therefore, the threshold frequency for the photoemission of electrons is $4.738 \times 10^{14} \mathrm{~Hz}$.