Question:
Two light waves of wavelengths 600 nm and 200 nm are incident on a metal surface. The maximum velocity of photoelectrons produced due to one wavelength is \(\frac{1}{3}\) of the maximum velocity of the photoelectrons produced due to the other wavelength. Find the work function of the metal.
Two light waves of wavelengths 600 nm and 200 nm are incident on a metal surface. The maximum velocity of photoelectrons produced due to one wavelength is \(\frac{1}{3}\) of the maximum velocity of the photoelectrons produced due to the other wavelength. Find the work function of the metal.
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For photoelectric effect with velocity ratio: relate \(v_{\text{max}}^2 \propto K_{\text{max}} = h\nu - \phi\) to solve for \(\phi\).
Updated On: Jul 18, 2026
- \(\frac{hc}{8} \times 10^7 \, \text{J}\)
- \(\frac{8}{hc} \times 10^7 \, \text{J}\)
- \(\frac{hc}{4} \times 10^7 \, \text{J}\)
- \(\frac{hc}{9} \times 10^7 \, \text{J}\)
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The Correct Option is D
Solution and Explanation
Step 1: Recall photoelectric effect equation.
\[ K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = h \nu - \phi \]
where \(\phi\) is work function, \(h\) Planck's constant, \(\nu = c/\lambda\).
Step 2: Express ratio of maximum velocities.
Given: \(v_1 = \frac{1}{3} v_2 \implies K_1 = \frac{1}{9} K_2\)
Step 3: Write kinetic energies in terms of wavelengths.
\[ K_1 = \frac{hc}{\lambda_1} - \phi, \quad K_2 = \frac{hc}{\lambda_2} - \phi \]
Step 4: Apply ratio.
\[ \frac{K_1}{K_2} = \frac{1}{9} \implies \frac{\frac{hc}{\lambda_1} - \phi}{\frac{hc}{\lambda_2} - \phi} = \frac{1}{9} \]
Step 5: Substitute \(\lambda_1 = 600 \, \text{nm}, \lambda_2 = 200 \, \text{nm}\).
\[ \frac{\frac{hc}{600} - \phi}{\frac{hc}{200} - \phi} = \frac{1}{9} \implies \phi = \frac{hc}{8} \]
Step 6: Final conclusion.
Hence, the work function of the metal is:
\[ \boxed{\frac{hc}{8} \times 10^7 \, \text{J}} \]
\[ K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = h \nu - \phi \]
where \(\phi\) is work function, \(h\) Planck's constant, \(\nu = c/\lambda\).
Step 2: Express ratio of maximum velocities.
Given: \(v_1 = \frac{1}{3} v_2 \implies K_1 = \frac{1}{9} K_2\)
Step 3: Write kinetic energies in terms of wavelengths.
\[ K_1 = \frac{hc}{\lambda_1} - \phi, \quad K_2 = \frac{hc}{\lambda_2} - \phi \]
Step 4: Apply ratio.
\[ \frac{K_1}{K_2} = \frac{1}{9} \implies \frac{\frac{hc}{\lambda_1} - \phi}{\frac{hc}{\lambda_2} - \phi} = \frac{1}{9} \]
Step 5: Substitute \(\lambda_1 = 600 \, \text{nm}, \lambda_2 = 200 \, \text{nm}\).
\[ \frac{\frac{hc}{600} - \phi}{\frac{hc}{200} - \phi} = \frac{1}{9} \implies \phi = \frac{hc}{8} \]
Step 6: Final conclusion.
Hence, the work function of the metal is:
\[ \boxed{\frac{hc}{8} \times 10^7 \, \text{J}} \]
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