Question

The wavelength of a \[1\,{\text{keV}}\] photon is \[1.24 \times {10^{ - 9}}\,{\text{m}}\]. What is the frequency of \[1\,{\text{MeV}}\] photon?
A. \[1.24 \times {10^{15}}\,{\text{Hz}}\]
B. \[2.4 \times {10^{20}}\,{\text{Hz}}\]
C. \[1.24 \times {10^{18}}\,{\text{Hz}}\]
D. \[2.4 \times {10^{23}}\,{\text{Hz}}\]

Answer 1

Hint: Use the formula for the energy of a photon in terms of the wavelength of the photon and the energy of a photon in terms of frequency of the photon. Convert the units of energies of the photon in the same unit. Substitute all the values in the two energy formulae and determine the required frequency.

Formulae used:
The energy \[E\] of a photon is given by
\[\Rightarrow E = \dfrac{{hc}}{\lambda }\] …… (1)
Here, \[h\] is the Planck’s constant, \[c\] is the speed of light and \[\lambda \] is the wavelength of the photon.
The energy \[E\] of a photon is given by
\[\Rightarrow E = h\gamma \] …… (2)
Here, \[h\] is the Planck’s constant and \[\gamma \] is the frequency of the photon.

Complete step by step solution:
We have given that the wavelength of the photon with energy \[1\,{\text{keV}}\] is \[1.24 \times {10^{ - 9}}\,{\text{m}}\].
\[\Rightarrow{E_1} = 1\,{\text{keV}}\]
\[\Rightarrow\lambda = 1.24 \times {10^{ - 9}}\,{\text{m}}\]
Convert the unit of energy \[{E_1}\] from keV to eV.
\[\Rightarrow{E_1} = \left( {1\,{\text{keV}}} \right)\left( {\dfrac{{{{10}^3}}}{{1\,{\text{k}}}}} \right)\]
\[ \Rightarrow {E_1} = 1 \times {10^3}\,{\text{eV}}\]
Hence, the energy \[{E_1}\] of the photon is \[1 \times {10^3}\,{\text{eV}}\].
Rewrite equation (1) for the energy \[{E_1}\] of the photon.
\[\Rightarrow{E_1} = \dfrac{{hc}}{\lambda }\]
Substitute \[1 \times {10^3}\,{\text{eV}}\] for \[{E_1}\] and \[1.24 \times {10^{ - 9}}\,{\text{m}}\] for \[\lambda \] in the above equation.
\[1 \times {10^3}\,{\text{eV}} = \dfrac{{hc}}{{1.24 \times {{10}^{ - 9}}\,{\text{m}}}}\] …… (3)
We need to determine the frequency of the photon with energy \[1\,{\text{MeV}}\].
\[\Rightarrow{E_2} = 1\,{\text{MeV}}\]
Convert the unit of energy \[{E_2}\] from MeV to eV.
\[\Rightarrow{E_2} = \left( {1\,{\text{MeV}}} \right)\left( {\dfrac{{{{10}^6}}}{{1\,{\text{M}}}}} \right)\]\[1 \times {10^6}\,{\text{eV}}\]
\[ \Rightarrow {E_2} = 1 \times {10^6}\,{\text{eV}}\]
Hence, the energy \[{E_2}\] of the photon is \[1 \times {10^6}\,{\text{eV}}\].
Rewrite equation (2) for the energy \[{E_2}\] of the photon.
\[\Rightarrow{E_2} = h\gamma \]
Substitute \[1 \times {10^6}\,{\text{eV}}\] for \[{E_2}\] in the above equation.
\[\Rightarrow 1 \times {10^6}\,{\text{eV}} = h\gamma \] …… (4)
Divide equation (4) by equation (3).
\[\Rightarrow\dfrac{{1 \times {{10}^6}\,{\text{eV}}}}{{1 \times {{10}^3}\,{\text{eV}}}} = \dfrac{{h\gamma }}{{\dfrac{{hc}}{{1.24 \times {{10}^{ - 9}}\,{\text{m}}}}}}\]
\[ \Rightarrow {10^3} = \dfrac{{\left( {1.24 \times {{10}^{ - 9}}\,{\text{m}}} \right)\gamma }}{c}\]
Rearrange the above equation for \[\gamma \].
\[ \Rightarrow \gamma = \dfrac{{c{{10}^3}}}{{1.24 \times {{10}^{ - 9}}\,{\text{m}}}}\]
Substitute \[3 \times {10^8}\,{\text{m/s}}\] for \[c\] in the above equation.
\[ \Rightarrow \gamma = \dfrac{{\left( {3 \times {{10}^8}\,{\text{m/s}}} \right){{10}^3}}}{{1.24 \times {{10}^{ - 9}}\,{\text{m}}}}\]
\[ \therefore \gamma = 2.4 \times {10^{20}}\,{\text{Hz}}\]
Therefore, the frequency of the photon is \[2.4 \times {10^{20}}\,{\text{Hz}}\].

Hence, the correct option is B.

Note: There is no need to convert the units of energy in the SI system of units as after taking division of the two energy equations, the units of energy term gets cancelled. But the students should not forget to convert the units of both the energies in the same unit like electronvolt for getting the correct answer.