The photo electric emission requires a threshold frequency $V_o$. For a certain metal $\lambda_1$ = 2200 $\overset{\circ}{A}$ and $\lambda_2$ = 1900 $\overset{\circ}{A}$ produce electrons with a maximum kinetic energy $KE_1$ and $KE_2$. If $KE_2 = 2KE_1$. Calculate $V_o$ and corresponding $\lambda_o$.

Question

$\begin{array}{1 1} (a)\;V_o= 1.1483S^{-1} \lambda = 26126\;\overset{\circ}{A}\\ (b)\;V_o= 1.1483\times10^{15}S^{-1} \lambda = 2612.6\;\overset{\circ}{A}\\ (c)\;V_o= 2.438\times10^{12}S^{-1} \lambda = 2634\;\overset{\circ}{A}\\ (d)\;V_o= 1\times10^{15}S^{-1} \lambda = 3642\;\overset{\circ}{A}\end{array}$