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)

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