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 problem involves the photoelectric effect, where ultraviolet (UV) light of energy is incident on a metal surface, causing electrons to be ejected. To find the maximum kinetic energy of these photoelectrons, we use the photoelectric equation given by Albert Einstein:

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

Where:

• \(K_{\text{max}}\) is the maximum kinetic energy of the photoelectrons.
• \(h \nu\) is the energy of the incident photon.
• \(\phi\) is the work function of the metal, the minimum energy required to eject an electron from the surface of the metal.

Given:

• Energy of the incident UV light, \(h \nu = 4.13 \text{ eV}\)
• Work function of the metal, \(\phi = 3.13 \text{ eV}\)

Substituting these values into the equation:

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

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

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

Why other options are incorrect:

• \(4.13 \text{ eV}\) is the energy of the incident light, not the kinetic energy.
• \(3.13 \text{ eV}\) is the work function, not the kinetic energy.
• \(7.26 \text{ eV}\) is obtained by incorrectly adding the incident light energy and the work function instead of subtracting.

The correct answer is indeed 1 eV.

Answer 2

The energy of the ejected photoelectrons is given by the photoelectric equation:

\( K.E. = h\nu - \phi \),

where \( h\nu \) is the energy of the incident photons and \( \phi \) is the work function of the material.

Given:

\( h\nu = 4.13 \, \text{eV}, \quad \phi = 3.13 \, \text{eV} \),

the maximum kinetic energy of the ejected photoelectrons is:

\( K.E. = 4.13 \, \text{eV} - 3.13 \, \text{eV} = 1 \, \text{eV} \).

Thus, the correct answer is Option (2).