Question

The work function of a certain metal is $3.31 \times 10^{-19} \,J$. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000\,\mathring{A}$ is
(Given, $h=6.62 \times 10^{-34}\, J - s , c =3 \times 10^{8} \,ms ^{-1}, \left.e=1.6 \times 10^{-19} \,C \right)$

• A. 2.48 eV
• B. 0.41 eV
• C. 2.07 eV
• D. 0.82 eV

Answer 1

Answer: (B)

Work function $W_{0}=3.31\times10^{-19} J$
Wavelength of incident radiation
$\lambda=5000\times10^{-10}\, m$
$E=W_{0}+KE$
According to Einstein's equation
$\Rightarrow \frac{hc}{\lambda}=3.31\times10^{-19}+KE$
$KE=-3.31\times10^{-19}+\frac{6.62\times10^{-34}\times3\times10^{8}}{5000\times10^{-10}}$
$=-3.31\times10^{-19}+\frac{6.62\times3}{5}\times10^{-19}$
$=\left(-3.31\times1.324\times3\right)\times10^{-19}$
$=\left(3.972-3.31\right)\times10^{-19}$
$=0.662\times10^{-19}\,J$
or $E=\frac{0.662\times10^{-19}}{1.6\times10^{-19}} $
$=0.41 \,eV$