Question

Energy of photon whose frequency is \[{10^{12}}MHz\], will be
A. \[4.14 \times {10^3}keV\]
B. \[4.14 \times {10^2}keV\]
C. \[4.14 \times {10^3}MeV\]
D. \[4.14 \times {10^3}eV\]

Answer 1

Hint: The energy carried out by a single photon is described as photon energy. This photon's energy is directly related to its electromagnetic frequency and inversely proportional to its wavelength. The higher the frequency of a photon, the greater its energy. The photon's energy is quantized, which means it cannot be split and is only available in discrete packets of energy.

Formula Used:
The formula for to find the energy of the photon is,
\[E = h\nu \]
Where, h is Planck’s constant and \[\nu \] is frequency.

Complete step by step solution:
Here we need to find the energy of the photon whose frequency is given as \[{10^{12}}MHz\]
Using the formula, we can write,
\[E = h\nu \]
Now, substitute the value of h and \[\nu \]in above equation, we get
\[E = 6.626 \times {10^{ - 34}} \times {10^{12}} \times {10^6}\]
\[\Rightarrow E = 6.626 \times {10^{ - 16}}J\]

In order to convert joules into eV we have to divide the obtained value by charge of an electron that is, \[1.6 \times {10^{ - 19}}\] then the above equation will become
\[E = \dfrac{{6.626 \times {{10}^{ - 16}}}}{{1.6 \times {{10}^{ - 19}}}}eV\]
\[\therefore E = 4.14 \times {10^3}eV\]
Therefore, the energy of the photon is \[4.14 \times {10^3}eV\].

Hence, Option D is the correct answer.

Note: The photon is the smallest packet (quanta) of energy, and according to the particle nature of light, light acts as particles, as proved by the photoelectric effect, while according to the wave nature of light, light behaves as waves, as confirmed by phenomena such as reflection and refraction. While solving this issue, keep in mind that if the energy is supplied in eV, we must convert the given value of energy from eV to joules by multiplying the charge of an electron.