Question

Radiation of wavelength $\lambda ,$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3\lambda }{4}$, the speed of the fastest emitted electron will be :

• A. $=v{\left(\frac{3}{4}\right)}^{1/2}$
• B. $<v{\left(\frac{4}{3}\right)}^{1/2}$
• C. $=v{\left(\frac{4}{3}\right)}^{1/2}$
• D. $>v{\left(\frac{4}{3}\right)}^{1/2}$

Answer 1

The correct option is D $>v{\left(\frac{4}{3}\right)}^{1/2}$

According to the Einstein's photoelectric equation,

$E=\varphi +K.E_{max}$

For case $\left(1\right)$,

$\frac{hc}{\lambda }=\varphi +\frac{1}{2}m{v}^2---\left(1\right)$

For case $\left(2\right)$,
$\frac{hc}{{\lambda }^{\prime }}=\varphi +\frac{1}{2}m(v^{\prime }{)}^2$

$\Rightarrow \frac{hc}{\left(\frac{3\lambda }{4}\right)}=\varphi +\frac{1}{2}m(v^{\prime }{)}^2---(2)$

$\left[\right(1)\times \frac{4}{3}]-\left(2\right)$

$\frac{4hc}{3\lambda }-\frac{4}{3}\frac{hc}{\lambda }=\frac{4}{3}\varphi +\frac{4}{3}\left(\frac{1}{2}m{v}^2\right)-\varphi -\frac{1}{2}m(v^{\prime }{)}^2$

$\Rightarrow \frac{4}{3}\varphi +\frac{4}{3}\left(\frac{1}{2}m{v}^2\right)=\varphi +\frac{1}{2}m(v^{\prime }{)}^2$

$\Rightarrow \frac{1}{2}m(v^{\prime }{)}^2=\frac{\varphi }{3}+\frac{4}{3}\frac{1}{2}m{v}^2$

$\Rightarrow \frac{1}{2}m(v^{\prime }{)}^2>\frac{4}{3}\left(\frac{1}{2}m{v}^2\right)$

$\Rightarrow v^{\prime }>\sqrt{\frac{4}{3}}v$

Hence, option $\left(D\right)$ is correct.