Question

The work function of a photosensitive metal surface is \(6.4 \times 10^{-19} \, {J}\). The maximum kinetic energy of the emitted photoelectrons when electromagnetic radiation of wavelength 1240 Å incidents on the metal surface is nearly:

• A. \(5 { eV}\)
• B. \(6 { eV}\)
• C. \(3 { eV}\)
• D. \(4 { eV}\)

Hint:

Ensure units are consistent when converting and applying physics formulas, especially in photoelectric effect calculations.

Answer 1

The Correct Option is B

To solve this problem, we use Einstein's photoelectric equation: \[ KE_{{max}} = E - \phi \] where \(KE_{{max}}\) is the maximum kinetic energy of the emitted photoelectrons, \(E\) is the energy of the incident photon, and \(\phi\) is the work function of the metal surface. First, we calculate the energy of the incident photon (\(E\)) using the wavelength \(\lambda = 1240 \, \AA\). 
We can use the approximation \(hc \approx 12400 \, {eV} \cdot \AA\). \[ E = \frac{hc}{\lambda} = \frac{12400 \, {eV} \cdot \AA}{1240 \, \AA} = 10 \, {eV} \] 
Next, we convert the work function \(\phi\) from Joules to electron volts (eV). We are given \(\phi = 6.4 \times 10^{-19} \, {J}\). 
We know that \(1 \, {eV} = 1.6 \times 10^{-19} \, {J}\). \[ \phi ({\text{in eV}}) = \frac{6.4 \times 10^{-19} \, {J}}{1.6 \times 10^{-19} \, {J/eV}} = 4 \, {eV} \] 
Now, we can find the maximum kinetic energy using Einstein's photoelectric equation: \[ KE_{{max}} = E - \phi = 10 \, {eV} - 4 \, {eV} = 6 \, {eV} \] The maximum kinetic energy of the emitted photoelectrons is nearly \(6 \, {eV}\). 
Correct Answer: (2) \(6 \, {eV}\)