Physics - Modern Physics Question with Solution | TestHub

Energy of the incident photon is $E=\frac{hc}{\lambda }$.

Since the work function is negligible, the kinetic energy of the emitted photoelectron is $K=E=\frac{hc}{\lambda }$.

The de-Broglie wavelength of the photoelectron is given by ${\lambda }_1=\frac{h}{p}=\frac{h}{\sqrt{2mK}}$.

Substituting the value of $K$, we get ${\lambda }_1=\frac{h}{\sqrt{2m\left(\frac{hc}{\lambda }\right)}}$.

Squaring both sides, ${\lambda }_1^2=\frac{h^2\lambda }{2mhc}=\frac{h\lambda }{2mc}$.

Therefore, $\lambda \propto {\lambda }_1^2$.

Answer: $\lambda \propto {\lambda }_1^2$