Question

In a photo-emissive cell, with exciting wavelength $\lambda$, the fastest electrons has speed $\mu$. If the exciting wavelength is changed to $3\lambda /4$, the speed of the fastest emitted electron will be

Answer 1

We know that,

$h\nu -{\omega }_o=\frac{1}{2}m{v}_{max}^2$

$=>\frac{hc}{\lambda }=\frac{hc}{{\lambda }_o}=\frac{1}{2}m{v}_{max}^2$

=> $hc\left(\frac{{\lambda }_o-\lambda }{\lambda {\lambda }_o}\right)=\frac{1}{2}m{v}_{max}^2$

$v_{max}=\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_o-\lambda }{\lambda {\lambda }_o}\right)}$---------- (1)

When wavelength is $\frac{3\lambda }{4}$ and velocity is v'

then $v^{/}=\sqrt{\frac{2hc}{m}\left[\frac{{\lambda }_o-\left(3\lambda /4\right)}{\left(3\lambda /4\right)\times {\lambda }_o}\right]}-------\left(2\right)$

Dividing equation, (2) by (1),

$\frac{v^{/}}{v}=\sqrt{\frac{[{\lambda }_o-(3\lambda /4\left)\right]}{\left(\frac{3}{4}\right)\lambda {\lambda }_o}\times \frac{\lambda {\lambda }_o}{{\lambda }_o-\lambda }}{v}^{/}=v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}\sqrt{\frac{[{\lambda }_o-(3\lambda /4\left)\right]}{{\lambda }_o-\lambda }}$

that is, $v^{/}=v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$