There are $${\eta _1}$$photon of frequency $${\gamma _1}$$in a beam of light. In an equally energetic beam, there are $${\eta _2}$$ photons of frequency $${\gamma _2}$$. Then find the correct relation.A. $$\dfrac{{{\eta _1}}}{{{\eta _2}}} = 1$$ B. $$\dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{{\gamma _1}}}{{{\gamma _2}}}$$C. $$\dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{{\gamma _2}}}{{{\gamma _1}}}$$D. $$\dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{{\gamma _1}^2}}{{{\gamma _2}^2}}$$

Question

There are $${\eta _1}$$photon of frequency $${\gamma _1}$$​in a beam of light. In an equally energetic beam, there are $${\eta _2}$$​ photons of frequency $${\gamma _2}$$​. Then find the correct relation.A. $$\dfrac{{{\eta _1}}}{{{\eta _2}}} = 1$$ B. $$\dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{{\gamma _1}}}{{{\gamma _2}}}$$C. $$\dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{{\gamma _2}}}{{{\gamma _1}}}$$D. $$\dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{{\gamma _1}^2}}{{{\gamma _2}^2}}$$

Vedantu Content Team · Accepted Answer

Hint: Photons are the very tiny particles that make up Electromagnetic waves. The energy and momentum of photons are dependent upon the wavelength and frequency of the electromagnetic wave. By the particle nature of light, we can say that the light behaves as particles and this is confirmed by the photoelectric effect, but according to the wave nature of light, light behaves as waves and is confirmed by phenomena like reflection, refraction, etc.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:Consider two beams of light of photons $${\eta _1}$$ and $${\eta _2}$$ having frequencies $${\gamma _1}$$ and $${\gamma _2}$$. The energy of these two beams is equal. We need to find the relation between these two. We know that,$$E = h\nu $$ If there are n number of photons then the above equation will become,$$E = \eta h\nu $$As they said that both the energies are equal, we can write as,$${E_1} = {E_2}$$ $$\Rightarrow {\eta _1}h{\gamma _1} = {\eta _2}h{\gamma _2}$$Here, they have given frequency as $$\gamma $$ instead of $$\nu $$. Then, $${\eta _1}{\gamma _1} = {\eta _2}{\gamma _2}$$$$ \therefore \dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{{\gamma _2}}}{{{\gamma _1}}}$$Therefore, the correct relation is $$\dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{{\gamma _2}}}{{{\gamma _1}}}$$Hence, Option C is the correct answerAdditional information: The energy of the photon is defined as the energy carried out by a single photon. The amount of energy of this photon is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. The higher the photon's frequency, the higher will be its energy.Note: Remember that basically to calculate the energy of the photon we use the formula, $$E = h\nu $$. In case we want to find the number of photons we write as, $$E = \eta h\nu $$ here, $$\eta $$ is the number of photons.