Question

The photoelectric threshold for a certain metal is $3600\, \mathring A$. The maximum energy of the ejected photoelectrons by radiation of $2000\, \mathring A$ is (Given $h = 6.62 \times 10^{-34} J \, s$ )

• A. 1.86 eV
• B. 2.76 eV
• C. 7.26 eV
• D. 5.76 eV

Answer 1

The Correct Option is B

Here,
Threshold wavelength, $\lambda_0 = 3600 $
$ \hspace55mm = 3600 \times 10^{-10} m $
$ \hspace55mm = 36 \times 10^{-8} m$
Incident wavelength, $\lambda = 2000 = 2000 \times 10^{-10} m $
$ \hspace55mm = 20 \times 10^{-8} m$
According to Einstein's photoelectric equation
$K_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda _{0}} = hc \left[\frac{1}{\lambda } - \frac{1}{\lambda_{0} }\right] $
$ = 6.62 \times 10^{-34} \times 3 \times 10^{8} \left[\frac{1}{20\times 10^{-8}} - \frac{1}{36 \times 10^{-8}}\right] $
$ = \frac{6.62 \times 10^{-34} \times 3\times 10^{8}}{10^{-8}} \left[\frac{1}{20} - \frac{1}{36}\right] $
$ = \frac{6.62 \times 10^{-34}\times 3 \times 10^{8}}{10^{-8}} \left[ \frac{9-5}{180}\right] $
$ = \frac{6.62 \times 10^{-34} \times 3\times 10^{8}\times 4}{180\times 10^{-8}} $
$ = \frac{6.62 \times 10^{-34}\times 3\times 10^{8}\times 4}{180 \times 10^{-8} \times 1.6 \times 10^{-19}} eV = 2.76 eV$