An important spectral emission line has a wavelength of 21 cm. The corresponding photon energy is
A. \[5.9 \times {10^{ - 4}}eV\]
B. \[5.9 \times {10^{ - 6}}eV\]
C. \[5.9 \times {10^{ - 8}}eV\]
D. \[11.8 \times {10^{ - 6}}eV\]

Hint:The photon is the qualitative unit of energy of the light wave. It is proportional to the frequency of the light wave. The frequency of the light wave is the most characteristic property because it doesn’t change by changing the medium in which the light is travelling.

Formula used:
\[E = h\nu \], where h is the Plank’s constant and E is the energy of the photon with frequency equal to \[\nu \]
\[c = \nu \lambda \], where c is the speed of light, \[\nu \]is the frequency of the photon and \[\lambda \]is the wavelength of the light wave.

Complete step by step solution:
The wavelength of the photon is given as 21 cm. The wavelength of the given photon in S.I. unit will be,
\[\lambda = 2.1 \times {10^{ - 1}}m\]
We got the wavelength of the photon and now we need to use the relation of the characteristic physical quantities for the electromagnetic wave to find the frequency of the photon,
\[c = \nu \lambda \]
\[\Rightarrow \nu = \dfrac{c}{\lambda }\]

On putting the value of the speed of light and the wavelength of the photon, we get
\[\nu = \dfrac{{3 \times {{10}^8}}}{{2.1 \times {{10}^{ - 1}}}}Hz\]
\[\Rightarrow \nu = 1.43 \times {10^9}Hz\]
Using the expression for the energy of the photon, we find,
\[E = h\nu \]
Putting the values of Plank’s constant and the frequency, we get
\[E = \left( {6.626 \times {{10}^{ - 34}}} \right) \times \left( {1.43 \times {{10}^9}} \right)J\]
\[\Rightarrow E = 9.47 \times {10^{ - 25}}J\]
We need to change the unit of energy from joule to eV.

One electron-Volt is the energy used to accelerate an electron in a region of electric potential 1 volt.
So, \[1eV = 1.6 \times {10^{ - 19}}J\]
Hence, the energy of the photon is,
\[E = \dfrac{{9.47 \times {{10}^{ - 25}}}}{{1.6 \times {{10}^{ - 19}}}}eV\]
\[\therefore E = 5.9 \times {10^{ - 6}}eV\]
Hence, the energy of the given photon is \[5.9 \times {10^{ - 6}}eV\].

Therefore, the correct option is B.

Note: While resolving numerical issues, we must be conscious about the physical quantity's units. All of the provided data must be converted into standard unit form. The photon's rest mass is also zero. A quantization of electromagnetic wave energy is the photon. The momentum of a particle is what determines motion.