Question

When incident wavelength is $\lambda$, stopping potential is $3V_0$. If incident wavelength is $2\lambda$ then stopping potential is $V_0$. Find out threshold wavelength in terms of $\lambda$.

Answer 1

Let ${\lambda }_0$ is the threshold wavelength. So, using relation as

$eV=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$

$3e{V}_0=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}).......\left(1\right)$

$e{V}_0=hc(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}).......\left(2\right)$

Dividing (1) by (2)

$3=\frac{(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})}{(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})}$

$3\times (\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})=(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$

$\frac{1}{2\lambda }=\frac{2}{{\lambda }_0}$

${\lambda }_0=4\lambda$

threshold wavelength in terms of ${\lambda }_0=4\lambda$