Question

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

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

Answer 1

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

We know,

$\frac{hc}{\lambda }-\varphi =\frac{1}{2}{mv}^2⟶\left(i\right)$

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

$\left(ii\right)$ - $\left(i\right)$ gives

$\frac{hc}{3\lambda }=\frac{1}{2}m(v^2-{v^{\prime }}^2)$

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

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