Question

Light of frequency $7.21\times 10^{14}\text{ Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0\times 10^5{\text{ m s}}^{-1}$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

• A. $4.74\times 10^{14}\text{ Hz}$
• B. $5.74\times 10^{15}\text{ Hz}$
• C. $6.84\times 10^{13}\text{ Hz}$
• D. $4.84\times 10^{14}\text{ Hz}$

Answer 1

Answer: (A)

Given, frequency of light, $\nu =7.21\times 10^{14}\text{Hz}$
Mass of electron, $m=9.1\times 10^{-31}\text{kg}$
Maximum speed of electrons, $v_{max}=6\times 10^5\text{ m/s}$
Let ${\nu }_0$ be the threshold frequency.
Use the formula for kinetic energy
$KE=\frac{1}{2}m{v}_{max}^2=h\nu -h{\nu }_0$
i.e., $\frac{1}{2}\times 9.1\times 10^{-31}\times 6\times 10^5\times 6\times 10^5=6.63\times 10^{-34}\left(v-v_0\right)$
or $v-v_0=\frac{36\times 9.1\times 10^{-21}}{2\times 6.63\times 10^{-34}}=2.47\times 10^{14}$
or $v_0=7.21\times 10^{14}-2.47\times 10^{14}\left(\because v=7.21\times 10^{14}Hz\right)$
$=4.74\times 10^{14}Hz$