Question

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

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

Answer 1

Answer: (C)

$\frac{1}{2}mv^{2}=\frac{hc}{\lambda }-\phi$ ...(i)
$\frac{1}{2}mv'^{2}=\frac{4}{3}\frac{hc}{\lambda }-\phi \, $ ...(ii)
From Eqn. (i)
$\frac{hc}{\lambda }=\frac{1}{2}mv^{2}+\phi$
On putting this Equation (ii)
$\frac{1}{2}\left(mv\right)^{′ 2}=\frac{4}{3}\left(\frac{1}{2} \left(mv\right)^{2} + \phi\right)-\phi$
$v'^{2}=\frac{4}{3}v^{2}+\frac{2 \phi}{3 m}$
$\Rightarrow v'>\sqrt{\frac{4}{3}}v$