Question

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

Answer 1

The correct answer is: $4.738×10^{14}Hz$
Frequency of the incident photon, $v=488nm=488×10^{-9}m$
Maximum speed of the electrons, $v=6.0×10^5m/s$
Planck’s constant, $h=6.626×10^{−34}Js$
Mass of an electron, $m=9.1×10^{−31}kg$
For threshold frequency $ν_0$, the relation for kinetic energy is written as:
$\frac{1}{2}mv^2=h(v-v_0)$
$v_0=v-\frac{mv^2}{2h}$
$=7.21×10^{14}-\frac{(9.1×10^{-31})×(6×10^5)^2}{2×(6.626×10^{-34})}$
$=7.21×10^{14}-2.472×10^{14}$
$=4.738×10^{14}Hz$
Therefore,the threshold frequency for the photoemission of electrons is $4.738×10^{14}Hz$