Question

A potential difference of \[2\,{\text{kV}}\] is applied across a X-ray tube and thus X-rays produced falls on a metallic plate at a potential of \[100\,{\text{V}}\]. What is the maximum KE of photoelectrons emitted metallic if its work function is \[5.8\,{\text{eV}}\].
A. \[2094.2\,{\text{eV}}\]
B. \[2105.8\,{\text{eV}}\]
C. \[1906.8\,{\text{eV}}\]
D. \[1894.2\,{\text{eV}}\]

Answer 1

Hint: Determine the net potential of X-rays incident on metal plate and use the formula for total energy incident on metal plate. Use the formula for total energy of the incident light on the metal plate in terms of the work function of the metal plate and maximum kinetic energy of the photoelectrons emitted.

Formulae used:
The expression for energy \[E\] of the incident ray in the photoelectric effect is given by
\[E = \phi + KE\] …… (1)
Here, \[\phi \] is the work function of the photoelectron and \[KE\] is the maximum kinetic energy of the photoelectrons.
The expression for energy \[E\] of the incident ray in the photoelectric effect is given by
\[E = eV\] …… (2)
Here, \[e\] is the charge on the photoelectron and \[V\] is the potential difference applied to the metallic plate.

Complete step by step answer:
We have given that a potential difference of \[2\,{\text{kV}}\] in applied across a X-ray tube because of which X-rays are produced in the tube and the potential of the metal plate is \[100\,{\text{V}}\]. Hence, the net potential in this process will be
\[V = \left( {2\,{\text{kV}}} \right) - \left( {100\,{\text{V}}} \right)\]
\[ \Rightarrow V = \left( {2000\,{\text{V}}} \right) - \left( {100\,{\text{V}}} \right)\]
\[ \Rightarrow V = 1900\,{\text{V}}\]
Hence, the net potential in this process is \[1900\,{\text{V}}\].

Let us now calculate the energy of the X-rays incident on the metallic plate.
Substitute \[1.6 \times {10^{ - 19}}\,{\text{C}}\] for \[e\] and \[1900\,{\text{V}}\] for \[V\] in equation (2).
\[E = \left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)\left( {1900\,{\text{V}}} \right)\]
\[ \Rightarrow E = 3040 \times {10^{ - 19}}\,{\text{J}}\]
Hence, the energy of the X-rays incident on the metal plate is \[3040 \times {10^{ - 19}}\,{\text{J}}\].

The work function of the metal plate is \[5.8\,{\text{eV}}\].
\[\phi = 5.8\,{\text{eV}}\]
Convert the unit of work function in the SI system of units.
\[\phi = \left( {5.8\,{\text{eV}}} \right)\left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)\]
\[ \Rightarrow \phi = 9.28 \times {10^{ - 19}}\,{\text{C}}\]
Hence, the value of work function of the metal plate is \[9.28 \times {10^{ - 19}}\,{\text{C}}\].

Let us now determine the maximum kinetic energy of the photoelectrons.
Rearrange equation (1) for maximum kinetic energy.
\[KE = E - \phi \]
Substitute \[3040 \times {10^{ - 19}}\,{\text{J}}\] for \[E\] and \[9.28 \times {10^{ - 19}}\,{\text{C}}\] for \[\phi \] in the above equation.
\[KE = \left( {3040 \times {{10}^{ - 19}}\,{\text{J}}} \right) - \left( {9.28 \times {{10}^{ - 19}}\,{\text{C}}} \right)\]
\[ \Rightarrow KE = 3030.72 \times {10^{ - 19}}\,{\text{J}}\]
Convert the unit of maximum kinetic energy in electronvolt.
\[ \Rightarrow KE = \left( {3030.72 \times {{10}^{ - 19}}\,{\text{J}}} \right)\left( {\dfrac{{1\,{\text{eV}}}}{{1.6 \times {{10}^{ - 19}}\,{\text{J}}}}} \right)\]
\[ \therefore KE = 1894.2\,{\text{eV}}\]
Therefore, the maximum kinetic energy of the photoelectrons emitted is \[1894.2\,{\text{eV}}\].

Hence, the correct option is D.

Note:The students should not forget to determine the net potential in this process of photoelectric effect by subtracting the potential of the metal plate from the potential applied to produce X-ray. If this is not done then the final answer for the maximum kinetic energy of photoelectrons will be incorrect.