A photosensitive metallic surface has work function $\phi$. If photon of energy $3 \phi$ falls on the…

A photosensitive metallic surface has work function $\phi$. If photon of energy $3 \phi$ falls on the surface, the electron comes out with a maximum velocity of $6 \times 10^6 \mathrm{~m} / \mathrm{s}$. When the photon energy is increased to $9 \phi$, then maximum velocity of photoelectrons will be
  1. $12 \times 10^6, \mathrm{~m} / \mathrm{s}$
  2. $6 \times 10^6 \mathrm{~m} / \mathrm{s}$
  3. $3 \times 10^6 \mathrm{~m} / \mathrm{s}$
  4. $24 \times 10^6 \mathrm{~m} / \mathrm{s}$

Solution

$\begin{aligned} & \text { For photoelèctric effect, } \\ & \mathrm{K} . \mathrm{E}_{\max }=\mathrm{E}-\phi_0 \\ & \text { Given: } E_1=3 \phi_0 \text { and } E_2=9 \phi_0 \\ & \text { From (i), } \\ & K . E_1=3 \phi_0-\phi_0=2 \phi_0 \\ & K . E_2=9 \phi_0-\phi_0=8 \phi_0 \text {. } \\ & \text { But, } K \cdot E_1=\frac{1}{2} \mathrm{mv}_1^2 \text { and } \mathrm{K} \cdot \mathrm{E}_2=\frac{1}{2} \mathrm{mv}_2^2 \\ & \therefore \quad \frac{\mathrm{~K} \cdot \mathrm{E}_1}{\mathrm{~K} \cdot \mathrm{E}_2}=\frac{\mathrm{v}_1^2}{\mathrm{v}_2^2}=\frac{1}{4} \\ & \therefore \quad \mathrm{v}_2=2 \mathrm{v}_1 \\ & \mathrm{v}_2=2 \times 6 \times 10^6=12 \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

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