Physics - Modern Physics Question with Solution | TestHub

According to Einstein's photoelectric equation, the stopping potential $V_0$ is related to the frequency $\nu$ of incident radiation by:

$e{V}_0=h\nu -\Phi$

where $\Phi$ is the work function of the material.

Rearranging the equation, we get:

$V_0=\left(\frac{h}{e}\right)\nu -\frac{\Phi }{e}$

This is the equation of a straight line. The intercept on the frequency axis (where $V_0=0$) represents the threshold frequency ${\nu }_0$. At this point:

$0=h{\nu }_0-\Phi \Rightarrow \Phi =h{\nu }_0$

From the given graph, the threshold frequency for surface A (${\nu }_{0A}$) is smaller than the threshold frequency for surface B (${\nu }_{0B}$), i.e., ${\nu }_{0A}<{\nu }_{0B}$.

Since the work function is directly proportional to the threshold frequency ($\Phi \propto {\nu }_0$), we have:

${\Phi }_A<{\Phi }_B$

Thus, the work function of A is smaller than that of B.

Answer: smaller than that of B.