Physics - Modern Physics Question with Solution | TestHub

According to Einstein's photoelectric equation, the stopping potential $V_s$ is related to the frequency $\nu$ of incident radiation by:

$e{V}_s=h\nu -\varphi$

$V_s=\frac{h}{e}\nu -\frac{\varphi }{e}$

The x-intercept of the $V_s$ versus $\nu$ graph represents the threshold frequency ${\nu }_0$, which occurs when $V_s=0$. Thus, ${\nu }_0=\frac{\varphi }{h}$, or $\varphi =h{\nu }_0$.

From the given graph, the threshold frequency for metal Y (${\nu }_0^{\prime }$) is greater than the threshold frequency for metal X (${\nu }_0$).

Since ${\nu }_0<{\nu }_0^{\prime }$, it follows that $h{\nu }_0<h{\nu }_0^{\prime }$.

Therefore, the work function of metal X is less than the work function of metal Y, i.e., ${\varphi }_x<{\varphi }_y$.

Answer: ${\varphi }_x<{\varphi }_y$