Question

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

• A. less than $v{(4/3)}^{1/2}$
• B. $v{(4/3)}^{1/2}$
• C. $v{(3/4)}^{1/2}$
• D. greater than $v{(4/3)}^{1/2}$

Answer 1

The correct option is D greater than $v{(4/3)}^{1/2}$

According to Einstein's photoelectric equation,
$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-W_{0}and\frac{1}{2}m{v_1}^2=\frac{hc}{\frac{3\lambda }{4}}-W_0=\frac{4}{3}\left(\frac{hc}{\lambda }-\frac{3}{4}{W}_0\right)=\frac{4}{3}\left(\frac{hc}{\lambda }-W_0+\frac{1}{4}{W}_0\right)=\frac{4}{3}\left(\frac{1}{2}m{v}^2+\frac{1}{4}{W}_0\right)\left(\right]$
So, $v_1$ is greater than $v{\left(\frac{4}{3}\right)}^{1/2}$.