Answer:
Option D
Explanation:
$k_{max}=\frac{hc}{\lambda_{ph}}-\phi$
Kinetic energy of electron reaching the anode will be
$K=\frac{h_{c}}{\lambda_{ph}}-\phi+eV$
Now $\lambda_{e}=\frac{h}{\sqrt{2mK}}$
$= \frac{h}{\sqrt{2m(\frac{hC}{\lambda_{ph}}-\phi+eV)}}$
If eV>>Φ
$\lambda_{e}= \frac{h}{\sqrt{2m(\frac{hC}{\lambda_{ph}}+eV)}}$
If $V_{f}=4V_{i}, (\lambda_{e)}f\approx (\lambda_{e)i}$