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

Answer 1

Answer: (A)

$\frac{hc}{\lambda }-\varphi =\frac{1}{2}m{v}^2$ .....(i)
$\frac{4hc}{3\lambda }-\varphi =\frac{1}{2}m{v}^{\prime 2}$ ....(ii)
$\frac{hc}{3\lambda }=\frac{1}{2}m\left(v^{\prime 2}-v^2\right)$
$\Rightarrow v^{\prime }=\sqrt{v^2+\frac{2hc}{3\lambda m}}$ ....(iii)
also from $\frac{hc}{\lambda }=\varphi +\frac{m{v}^2}{2}$
$\Rightarrow \frac{2hc}{\lambda m}=\frac{2\varphi }{m}+v^2$ /.
$\Rightarrow \frac{2hc}{3\lambda m}=\frac{2\varphi }{3m}+\frac{r^2}{3}>\frac{v^2}{3}$ .....(iv)
combining (iii) & (iv)
$v^{\prime }>\sqrt{\frac{4v^2}{3}}$