Question

The ratio of the energy of a photon with \[\lambda = 150\,nm\] to that with \[\lambda = 300\,nm\] is
A. 2
B. \[\dfrac{1}{4}\]
C. 4
D. \[\dfrac{1}{2}\]

Answer 1

Hint: The photon is the basic unit of the energy of the electromagnetic wave. In order to solve this problem we have to use the expression of energy of the photon and relation between the wavelength of the light and its frequency.

Formula used:
\[c = \nu \lambda \],
Here \[\nu \] is the frequency of the electromagnetic wave and \[\lambda \] is the wavelength of the same electromagnetic wave in vacuum.
\[E = h\nu \]
Here E is the energy of the photon, h is the Plank’s constant and \[\nu \] is the frequency of the electromagnetic wave.

Complete step by step solution:
The wavelengths of two photons are given as 150 nm and 300 nm
\[{\lambda _1} = 150\,nm\]
\[\Rightarrow {\lambda _2} = 3000\,nm\]
Then the frequencies of the two photons will be determined using the relation between the speed of the electromagnetic wave, frequency and wavelength.

The speed of electromagnetic waves is constant in a particular medium of propagation. So the frequency of the wave is inversely proportional to the wavelength.
\[c = \nu \lambda \]
\[\Rightarrow \nu = \dfrac{c}{\lambda }\]
\[\Rightarrow \nu \propto \dfrac{1}{\lambda }\]
Hence, the ratio of the frequencies of the photon is,
\[\dfrac{{{\nu _1}}}{{{\nu _2}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}\]
\[\Rightarrow \dfrac{{{\nu _1}}}{{{\nu _2}}} = \dfrac{{300\,nm}}{{150\,nm}}\]
\[\Rightarrow \dfrac{{{\nu _1}}}{{{\nu _2}}} = 2 \ldots \left( i \right)\]

Using the formula of the energy of the photon,
\[E = h\nu \]
Here, E is the energy of the photon and h is the Plank’s constant. Hence, the energy is directly proportional to the frequency of the photon.
\[E \propto \dfrac{1}{\nu }\]
So, the ratio of the energies of the photons is,
\[\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{\nu _1}}}{{{\nu _2}}} \ldots \left( {ii} \right)\]
From the first and second relation, we get the ratio of the energies of the photons as,
\[\dfrac{{{E_1}}}{{{E_2}}} = 2\]
Hence, the ratio of the energies of the two photons is 2.

Therefore, the correct option is A.

Note: We must be careful about the medium of the electromagnetic wave in which it is propagating. The wavelength changes as we change the medium. The wavelength of the electromagnetic wave is inversely proportional to the refractive index of the medium.