Two photons having energies twice and thrice the work function of metal are incident one after another on…
Two photons having energies twice and thrice the work function of metal are incident one after another on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases is respectively
$\sqrt{3}: 3$
$\sqrt{2}: \sqrt{3}$
$1: \sqrt{2}$
$\sqrt{3}: 1$
Solution
$\begin{aligned} & K \cdot E_{\max }=h v-\phi_0 \\ & \text {Given: } E_1=2 \phi_0 \text { and } E_2=3 \phi_0 \\ & \Rightarrow K \cdot E_1=2 \phi_0-\phi_0=\phi_0 \\ & \Rightarrow K \cdot E_2=3 \phi_0-\phi_0=2 \phi_0 \\ & \text {but, } K \cdot E_1=\frac{1}{2} m v_1^2 \text { and } K \cdot E_2=\frac{1}{2} m v_2^2 \\ \therefore \quad & \frac{K \cdot E_1}{K \cdot E_2}=\frac{V_1^2}{V_2^2}=\frac{1}{2} \\ \therefore \quad & \frac{V_1}{V_2}=\frac{1}{\sqrt{2}}\end{aligned}$
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