The maximum wavelength of light which causes photoelectric emission from a photosensitive metal surface is…

The maximum wavelength of light which causes photoelectric emission from a photosensitive metal surface is $\lambda_0$. Two light beams of wavelengths $\frac{\lambda_0}{3}$ and $\frac{\lambda_0}{9}$ incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is
  1. $3: 4$
  2. $1: 3$
  3. $1: 2$
  4. $2: 3$

Solution

By Einstein's photoelectric equation, $\begin{aligned} & \mathrm{E}=\phi_{\mathrm{o}}+\mathrm{K}_{\max } \\ & \Rightarrow \frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}=\frac{1}{2} \mathrm{mv}_{\max }^2 \end{aligned}$
$\begin{aligned} & \Rightarrow v_{\max }^2=\frac{2 \mathrm{mc}}{\mathrm{m}}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{\mathrm{o}}}\right) \\ & \Rightarrow \frac{\mathrm{v}_{\max }^2, 1}{\mathrm{v}_{\max , 2}^2}=\frac{\left(\frac{1}{\lambda_0 / 3}-\frac{1}{\lambda_0}\right)}{\left(\frac{1}{\lambda_0 / 9}-\frac{1}{\lambda_0}\right)}=\frac{\frac{2}{\lambda_0}}{\frac{8}{\lambda_0}}=\frac{1}{4} \\ & \therefore \frac{V_{\max , 1}}{V_{\max , 2}}=\frac{1}{2}\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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