Question

The retarding potential necessary to stop the emission of photoelectrons, when a target material of work function $1.24\, eV$ is irradiated with light of wavelength $4.36 \times 10^{7} m$ is

• A. $4.08\, eV$
• B. $2.84\, eV$
• C. $1.60\, eV$
• D. $0.36\, eV$

Answer 1

Answer: (C)

Given, work function, $\varphi_{0}=1.24\, eV$
$=1.24 \times 1.6 \times 10^{-19} J$
Wavelength, $\lambda=4.36 \times 10^{-7} m$
According to Einstein's photoelectric equation,
$e V_{0}=\frac{h c}{\lambda}-\varphi$
$\Rightarrow e V_{0}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{4.36 \times 10^{-7}}-1.24 \times 1.6 \times 10^{-19}$
$V_{0}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 4.36 \times 10^{-7}}-1.24$
$=2.85-1.24=1.61\, V \sim e q 1.60\, V$