Question

In a photocell, with excitation wavelength $\lambda$, the faster electrons has speed $v$. If the excitation wavelength is changed to $3\lambda /4$, the speed of the fastest electron will be

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

Answer 1

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

Initially, using energy conservation
$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-W_0$
Let the speed of the fastest electron be $v_1,$ when excitation wavelength is changed to $3\lambda /4$.
$\therefore \frac{1}{2}m{v}_1^2=\frac{4hc}{3\lambda }-W_0$
$\Rightarrow \frac{1}{2}m{v}_1^2=\frac{4}{3}(\frac{hc}{\lambda }-W_0)+\frac{W_0}{3}$
$\Rightarrow \frac{1}{2}m{v}_1^2=\frac{4}{3}\left(\frac{1}{2}m{v}^2\right)+\frac{W_0}{3}\left[\text{using Eq. (i)}\right]$
$\Rightarrow v_1^2=\frac{4v^2}{3}+\frac{2W_0}{3m}$
$\therefore v_1>\sqrt{\frac{4}{3}}v$