Question

Ratio of energy of photon of wavelength \[3000{\text{ }}{{\text{A}}^ \circ }\] and \[6000{\text{ }}{{\text{A}}^ \circ }\]
A. \[3:1\]
B. \[2:1\]
C. \[1:2\]
D. \[1:3\]

Answer 1

Hint: In this question, we are given wavelengths of two photons, and we need to find the ratio of the energy of photons. As we know, from Planck’s quantum theory of radiation, light travels in the form of discrete packets called photons, and the energy of photons is given by \[E = hv\] and \[c = v\lambda \]. Using these relations, we can find out the ratio of the energy of photons.

Formula used:
Energy of a photon, \[E = hv\]
Where \[h = 6.63 \times {10^{ - 34}}{\text{ Js}}\] (Planck’s constant)
\[v\] is the frequency of photon

Complete step by step answer:
We have given,
Wavelength of the first photon is \[{\lambda _1} = 3000{\text{ }}{{\text{A}}^ \circ }\]
Wavelength of the second photon is \[{\lambda _2} = 6000{\text{ }}{{\text{A}}^ \circ }\]
According to Planck’s quantum theory of radiation light travels in form of discrete packet called photons and energy of photons which travels with the speed of light
Thus, energy of photon is given by, \[E = hv\] where \[h\] is the Planck’s constant
But we know \[c = v\lambda \]
\[ \Rightarrow v = \dfrac{c}{\lambda }\]
Substituting this in the equation \[E = hv\] we get,
\[E = \dfrac{{hc}}{\lambda }\]
Energy of the first photon \[{E_1} = \dfrac{{hc}}{{{\lambda _1}}}\]
Energy of the second photon \[{E_2} = \dfrac{{hc}}{{{\lambda _2}}}\]
Energy ratio is given by \[\dfrac{{{E_1}}}{{{E_2}}}\]
Thus, \[\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{\dfrac{{hc}}{{{\lambda _1}}}}}{{\dfrac{{hc}}{{{\lambda _2}}}}}\]
On simplifying we get,
\[ \Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}\]
Substituting \[{\lambda _1} = 3000{\text{ }}{{\text{A}}^ \circ }\] and \[{\lambda _2} = 6000{\text{ }}{{\text{A}}^ \circ }\] we get,
\[ \Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{6000}}{{3000}}\]
\[{E_1}:{E_2} = 2:1\]
Thus, ratio of energy of photon is \[2:1\]
Hence, the correct option is option B.

Additional information:
 The rest mass of the photon is zero,which means it cannot exist at rest. According to the theory of relativity, the mass \[m\] of a particle moving with velocity \[{\text{v}}\], comparable with velocity of light \[{\text{c}}\] is given by \[{\text{m = }}\dfrac{{{m_o}}}{{\sqrt {\left( {1 - \dfrac{{{v^2}}}{{{c^2}}}} \right)} }}\] where \[{m_o}\] is the rest mass of particle. As a photon moves with a speed of light \[{m_o}\] becomes zero. So, the rest mass of the photon is zero.

Note: Irrespective of the intensity of radiation, all the photons of a particular frequency or wavelength have the same energy. As the energy of a photon depends upon the frequency, the energy of the photon does not change when the photon travels from one medium to another.