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