Question

A photon of energy 8eV is incident on a metal surface of threshold frequency \[1.6 \times {10^{15}}Hz\]. The maximum kinetic energy of the photoelectrons emitted (in eV) (Take \[h = 6.6 \times {10^{ - 34}}Js\]).
A. 1.4 eV
B. 2.4 eV
C. 4.8 eV
D. 0.8 eV

Answer 1

Hint: The maximum kinetic energy of the photoelectron is the energy carried by the electron after getting out of the metal surface. A part of the photon energy is used to eject the electron and rest of the energy is transferred to the ejected electron in the form of kinetic energy.

Formula used:
\[K = h\nu - \phi \]
where K is the maximum kinetic energy of the photoelectron, \[\nu \] is the frequency of the photon incident on the metal surface of the work function \[\phi \].

Complete step by step solution:
The energy of the photon is given as 8 eV. The threshold frequency of the surface of the metal is given as \[1.6 \times {10^{15}}Hz\]. We need to find the maximum kinetic energy of the photoelectrons emitted (in eV).Using the threshold frequency of the metal surface, the minimum energy for the photoelectrons to eject from the surface of the metal will be the work function of the metal surface.
\[\phi = h{\nu _0}\]
\[\Rightarrow \phi = \left( {6.6 \times {{10}^{ - 34}}} \right) \times \left( {1.6 \times {{10}^{15}}} \right)J\]
\[\Rightarrow \phi = 1.06 \times {10^{ - 18}}J\]

As the calculated work function of the metal surface is in the unit of joule and the energy of the photon is in eV, we convert the work function to the unit of eV.
As we know that \[1eV = 1.6 \times {10^{ - 19}}J\]
\[\phi = \dfrac{{1.06 \times {{10}^{ - 18}}}}{{1.6 \times {{10}^{ - 19}}}}eV\]
\[\Rightarrow \phi = 6.6eV\]
So, the work function of the metal surface is 6.6 eV and the energy of the photon is 8 eV.

The maximum kinetic energy of the photoelectron from the formula is the difference between the energy of the photon and the work function. So, using the formula of maximum kinetic energy, we get
\[K = 8.0eV - 6.6\,eV\]
\[\therefore K = 1.4\,eV\]
Hence, the maximum kinetic energy of the photoelectrons is 1.4 eV.

Therefore, the correct option is A.

Note: As all the units of energies are given in electron-volts and the final answer also in electron-volts, so we don’t need to convert the given energy into standard units before putting in the formula to get the final answer.