Question:

If the wavelength of the incident radiation on a photosensitive metal surface is decreased from \(3100 \, \text{Å}\) to \(1550 \, \text{Å}\), the maximum kinetic energy of the emitted photoelectrons is tripled. - The work function of the metal surface is

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For photoelectric effect problems, use Einstein’s equation: \[ h f = \phi + K_{\text{max}} \] where \( E = \frac{hc}{\lambda} \). Ensure wavelengths are converted to consistent units.
Updated On: May 5, 2026
  • \(3 \, eV\)
  • \(4 \, eV\)
  • \(2 \, eV\)
  • \(6 \, eV\)
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The Correct Option is C

Solution and Explanation

Step 1: Einstein’s Photoelectric Equation

\[ hf = \phi + K_{\max} \]

Photon energy:

\[ E = \frac{hc}{\lambda}, \quad hc = 12400 \, \text{eV·\AA} \]

Step 2: Calculate photon energies

For \( \lambda_1 = 3100 \, \text{\AA} \):

\[ E_1 = \frac{12400}{3100} = 4 \, \text{eV} \]

For \( \lambda_2 = 1550 \, \text{\AA} \):

\[ E_2 = \frac{12400}{1550} = 8 \, \text{eV} \]

Step 3: Apply given condition

\[ E_1 = \phi + K_1 \] \[ E_2 = \phi + 3K_1 \]

Subtracting:

\[ E_2 - E_1 = 2K_1 \]

\[ 8 - 4 = 2K_1 \Rightarrow K_1 = 2 \, \text{eV} \]

Step 4: Find work function

\[ \phi = E_1 - K_1 = 4 - 2 = 2 \, \text{eV} \]

Final Answer: \( 2 \, \text{eV} \)

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