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
$K E = \frac{1}{2} m v_{max}^{2} = h \nu - h \nu_{0}$
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 $