Question

A light beam of wavelength \[400\,{\text{nm}}\] is incident on a metal plate of work function \[2.2\,{\text{eV}}\]. A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal.

Answer 1

Hint: Use the formula for energy of a photon in terms of the wavelength of the photon. Also use the formula for total energy of the photon in the photoelectric effect in terms of work function of the metal and the kinetic energy of the electron. Calculate the energy lost by the electron in the two collisions and subtract it from the energy of the incident light ray.

Formulae used:
The energy \[E\] of a photon is
\[E = \dfrac{{hc}}{\lambda }\] …… (1)
Here, \[h\] is Planck’s constant, \[c\] is speed of light and \[\lambda \] is wavelength of the light.
The total energy \[E\] of the photon in the photoelectric effect is
\[E = \phi + K\] …… (2)
Here, \[\phi \] is the work function of the metal and \[K\] is the kinetic energy of the electron.

Complete step by step answer:
We have given that a light beam of the wavelength \[400\,{\text{nm}}\] is incident on the metal plate and the work function of the metal plate is \[2.2\,{\text{eV}}\].
\[\lambda = 400\,{\text{nm}}\]
\[\phi = 2.2\,{\text{eV}}\]
We have asked to calculate the kinetic energy of the electron coming out of the metal.

Let us first calculate the energy of the incident light beam.Substitute \[1240\,{\text{eV}} \cdot {\text{nm}}\] for \[hc\] and \[400\,{\text{nm}}\] for \[\lambda \] in equation (1).
\[E = \dfrac{{1240\,{\text{eV}} \cdot {\text{nm}}}}{{400\,{\text{nm}}}}\]
\[ \Rightarrow E = 3.1\,{\text{eV}}\]
Hence, the energy of the light beam incident of the metal plate is \[3.1\,{\text{eV}}\].

We have given that the electron ejected from the metal plate undergoes collision two times before coming out of the metal plate.The energy lost in the first collision is
\[\Delta {E_1} = \left( {3.1\,{\text{eV}}} \right) \times \dfrac{{10}}{{100}}\]
\[ \Rightarrow \Delta {E_1} = 0.31\,{\text{eV}}\]
Hence, the energy of the electron after first collision is
\[ \Rightarrow {E_1} = 3.1\,{\text{eV}} - 0.31\,{\text{eV}}\]
\[ \Rightarrow {E_1} = {\text{2}}{\text{.79}}\,{\text{eV}}\]
Hence, the energy of the electron after the first collision is \[{\text{2}}{\text{.79}}\,{\text{eV}}\].

The energy lost by the electron in the second collision is
\[\Delta {E_2} = \left( {{\text{2}}{\text{.79}}\,{\text{eV}}} \right) \times \dfrac{{10}}{{100}}\]
\[ \Rightarrow \Delta {E_2} = 0.279\,{\text{eV}}\]
The total energy lost by the electron during the collision is
\[\Delta E = \Delta {E_1} + \Delta {E_2}\]
\[ \Rightarrow \Delta E = \left( {0.31\,{\text{eV}}} \right) + \left( {0.279\,{\text{eV}}} \right)\]
\[ \Rightarrow \Delta E = 0.589\,{\text{eV}}\]
Hence, the total energy lost by the electron during two collisions is \[0.589\,{\text{eV}}\].

Let us now calculate the kinetic energy of the electron coming out of the metal plate using equation (2).According to equation (2), the kinetic energy of the electron is
\[K = E - \phi - \Delta E\]
Substitute \[3.1\,{\text{eV}}\] for \[E\], \[2.2\,{\text{eV}}\] for \[\phi \] and \[0.589\,{\text{eV}}\] for \[\Delta E\] in the above equation.
\[K = \left( {3.1\,{\text{eV}}} \right) - \left( {2.2\,{\text{eV}}} \right) - \left( {0.589\,{\text{eV}}} \right)\]
\[ \therefore K = 0.311\,{\text{eV}}\]

Hence, the kinetic energy of the electron coming out of the metal plate is \[0.311\,{\text{eV}}\].

Note:The students should not forget to calculate the total energy of the incident light beam on the metal plate in the SI system of units. Also the student should not forget to subtract the total energy lost by the electron in two collisions from the energy of the incident beam of light. If this energy is not subtracted then the final answer for the kinetic energy will be incorrect.