JEE Advanced · 2006 · 09 Apr 2006 · Modern Physics

From Einstein's photoelectric equation: $\frac{hc}{\lambda }-\varphi =eV$, which gives $V=\frac{hc}{e\lambda }-\frac{\varphi }{e}$ or $V=\frac{hc}{e}\cdot \frac{1}{\lambda }-\frac{\varphi }{e}$.

This is a linear relation between V and $1/\lambda$ with slope $\tan \theta =\frac{hc}{e}$ (same for all metals) and y-intercept $-\frac{\varphi }{e}$.

From the graph, the x-intercepts (where V=0) give $\frac{1}{{\lambda }_0}=\frac{\varphi }{hc}$. Reading from the graph: for metal 1, $\frac{{\varphi }_1}{hc}=0.001$; for metal 2, $\frac{{\varphi }_2}{hc}=0.002$; for metal 3, $\frac{{\varphi }_3}{hc}=0.004$. Therefore ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$ (option A correct, B incorrect).

The slope $\tan \theta =\frac{hc}{e}$ is directly proportional to hc/e (option C correct).

For metal 2: ${\lambda }_0=\frac{hc}{{\varphi }_2}=\frac{1}{0.002}=500$ nm. For metal 3: ${\lambda }_0=\frac{1}{0.004}=250$ nm. Violet light has $\lambda \approx 400$ nm. Since ${\lambda }_{violet}<{\lambda }_0$ for metal 2 (400 < 500), photoelectrons will be ejected from metal 2. But ${\lambda }_{violet}>{\lambda }_0$ for metal 3 (400 > 250), so no photoelectrons from metal 3 (option D incorrect).