Question

In a photoelectric effect experiment, the maximum kinetic energy of the ejected photoelectrons is measured for various wavelengths of the incident light. Figure shown a graph of this maximum kinetic energy $K_{max}$ as a function of the wavelength $\lambda$ of the light falling on the surface of the metal. Which of the following statement/s is/are correct?

• A. Threshold frequency for the metal is $1.2\times {10}^{15}m$
• B. Work function of the metal is 4.968 eV
• C. Maximum kinetic energy of photoelectrons corresponding to light of wavelength 100 nm is nearly 7.4 eV
• D. Photoelectric effect takes place with red light

Answer 1

Answer: (A), (B), (C)

The correct options are
A Threshold frequency for the metal is $1.2\times {10}^{15}m$
B Work function of the metal is 4.968 eV
C Maximum kinetic energy of photoelectrons corresponding to light of wavelength 100 nm is nearly 7.4 eV

From figure, $K_{max}=0$ $@$ $\lambda =250$ nm

$\therefore$ Threshold frequency ${\nu }_{th}=\frac{3\times {10}^8}{250\times {10}^{-9}}=1.2\times {10}^{15}$ Hz

Photoelectric effect takes place only with the incident light having frequency greater than threshold frequency. As frequency of red light is less than ${\nu }_{th}$, hence photoelectric effect does not take place with red light.

Work function $\varphi =\frac{h{\nu }_{th}}{e}=\frac{6.6\times {10}^{-34}\times 1.2\times {10}^{15}}{1.6\times {10}^{-19}}=4.968$ eV

Using $K_{max}=\frac{hc}{\lambda }-\varphi$

Given : $\lambda =100$ nm

$\therefore$ $K_{max}=\frac{1242.4}{100}-4.968$ $⟹K_{max}=7.4$ eV