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

• A. \(3 \, eV\)
• B. \(4 \, eV\)
• C. \(2 \, eV\)
• D. \(6 \, eV\)

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Using Einstein’s Photoelectric Equation The photoelectric equation is given by: \[ h f = \phi + K_{\text{max}} \] where: 
- \( h f \) is the incident photon energy, 
- \( \phi \) is the work function, 
- \( K_{\text{max}} \) is the maximum kinetic energy of the emitted electrons. Photon energy is related to wavelength by: \[ E = \frac{hc}{\lambda} \] where: 
- \( h c = 12400 \) eV·Å, 
- \( \lambda \) is the wavelength in Å. 

Step 2: Calculating Photon Energies For \(\lambda_1 = 3100\) Å: \[ E_1 = \frac{12400}{3100} = 4 \, eV \] For \(\lambda_2 = 1550\) Å: \[ E_2 = \frac{12400}{1550} = 8 \, eV \] 

Step 3: Using the Given Condition Let the initial kinetic energy be \( K_1 \), and the final kinetic energy be \( K_2 = 3K_1 \). Using the photoelectric equation: \[ E_1 = \phi + K_1 \] \[ E_2 = \phi + 3K_1 \] Substituting values: \[ 4 = \phi + K_1 \] \[ 8 = \phi + 3K_1 \] Solving for \( \phi \): \[ (8 - 4) = (3K_1 - K_1) \] \[ 4 = 2K_1 \] \[ K_1 = 2 \, eV \] Substituting \( K_1 \) in \( \phi = 4 - K_1 \): \[ \phi = 4 - 2 = 2 \, eV \] Thus, the correct answer is \( \mathbf{(3)} \ 2 \, eV \).