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
- 4.13 eV
- 1 eV
- 3.13 eV
- 7.26 eV
The Correct Option is B
Approach Solution - 1
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.
Approach Solution -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).
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