IIT JEE Advanced 2006 Paper 1

From Einstein’s photoelectric equation, we get $\frac{hc}{\lambda}$ = ϕ + $\mathrm{eV}_0$ where ϕ is the work function and $V_0$ is the stopping potential: $V_0$ = $\frac{hc}{\lambda c} - \frac{\phi}{e}$ On comparing with y = mx + c, we get the slope m = $\frac{hc}{e}$ = tan θ • For metal (1): $\frac{1}{\lambda_{01}}$ = $\frac{\phi_1}{hc}$ = 0.001 • For metal (2): $\frac{1}{\lambda_{02}}$ = $\frac{\phi_2}{hc}$ = 0.002 • For metal (3): $\frac{1}{\lambda_{03}}$ = $\frac{\phi_3}{hc}$ = 0.004 Therefore, $\phi_1:\phi_2\text{''}\phi_3$ = 1 : 2 : 4 For metal (2), the threshold is $\lambda_{02}$ = $\frac{hc}{\phi_2}$ = $\frac{1000}{2}$ = 500 nm For metal (3), the threshold is $\lambda_{03}$ = $\frac{hc}{\phi_3}$ = $\frac{1000}{4}$ = 250 nm (a) $\lambda_{\text{violet}}$ = 400 nm < $\lambda_{02}$. (b) $\lambda_{\text{violet}}$ ejects photoelectrons from metal (2). (c) $\lambda_{\text{violet}}$ = 400 nm < $\lambda_{03}$. (d) $\lambda_{\text{violet}}$ cannot eject photoelectrons from metal (3).