Question

UV light of 4.13 eV is incident on a photosensitive metal surface having work function 3.13 eV. The maximum kinetic energy of ejected photoelectrons will be

• A. 4.13 eV
• B. 1 eV
• C. 3.13 eV
• D. 7.26 eV

Answer 1

The Correct Option is B

The photoelectric effect describes the ejection of electrons from a metal surface when illuminated by ultraviolet (UV) light of a specific energy. The maximum kinetic energy of these photoelectrons is determined by Albert Einstein's photoelectric equation:

\(K_{\text{max}} = h u - \phi\)

Where:

• \(K_{\text{max}}\) represents the maximum kinetic energy of the photoelectrons.
• \(h u\) is the energy of the incident photon.
• \(\phi\) denotes the work function of the metal, which is the minimum energy necessary to liberate an electron from the metal's surface.

Provided data:

• Incident UV light energy: \(h u = 4.13 \text{ eV}\)
• Metal's work function: \(\phi = 3.13 \text{ eV}\)

Applying these values to the equation:

\(K_{\text{max}} = 4.13 \text{ eV} - 3.13 \text{ eV}\)

\(K_{\text{max}} = 1 \text{ eV}\)

Therefore, the maximum kinetic energy of the ejected photoelectrons is 1 eV.

Evaluation of alternative options:

• \(4.13 \text{ eV}\) corresponds to the incident light energy, not the kinetic energy.
• \(3.13 \text{ eV}\) represents the work function, not the kinetic energy.
• \(7.26 \text{ eV}\) results from the erroneous addition of incident light energy and work function, rather than their subtraction.

The accurate result is confirmed to be 1 eV.