Question

A green bulb and a red bulb are emitting the radiations with equal power. The correct relation between numbers of photons emitted by the bulbs per second is:
A) ${{n}_{g}}={{n}_{r}}$
B) ${{n}_{g}}<{{n}_{r}}$
C) ${{n}_{g}}>{{n}_{r}}$

Answer 1

Hint:. The answer to this question is based on the concept of energy of photon where it is inversely proportional to wavelength which is given by the formula$E=\dfrac{hc}{\lambda }$

Complete step by step answer:
Photon energy basically refers to the energy carried by a single photon. We have come across this concept in radiation chemistry and also in physics as well.
- Photon energy can be expressed in any form of unit of energy like eV or joules etc.
- According to the concept of photon energy, the energy of a photon is inversely proportional to the wavelength of photon and is directly proportional to the electromagnetic radiation of the photons.
- Therefore, higher the frequency of the photon, higher will be the energy of the photon. Also longer the wavelength of the photon, lower will be the energy of the photon.

- Based on these facts we can write an equation of the energy of photon as,
$E=\dfrac{hc}{\lambda }$ ……..(1)
where e is the energy of photon
h is the constant called as Planck’s constant which has the value $6.626\times {{10}^{-34}}Js$
c is the velocity of light which is a constant and has value $3\times {{10}^{8}}m/s$
and $\lambda $ is the wavelength of photon

Thus , if we denote number of photons emitted by the green and red bulb as ${{n}_{g}}$ and ${{n}_{r}}$ respectively then according to equation (1),
$\dfrac{{{n}_{g}}(hc)}{{{\lambda }_{g}}}=\dfrac{{{n}_{r}}(hc)}{{{\lambda }_{r}}}$
\[\Rightarrow {{n}_{g}}=\dfrac{{{\lambda }_{g}}\times {{n}_{r}}}{{{\lambda }_{r}}}\]
Since, ${{\lambda }_{g}}<{{\lambda }_{r}}$ according to absorption spectrum, we have ${{n}_{g}}<{{n}_{r}}$
So, the correct answer is “Option B”.

Note: Important point to be noted is that the photon energy and De-Broglie wavelength have similar formulas where $\lambda =\dfrac{h}{mv}$ for De-Broglie wavelength and for photon energy it is $\lambda =\dfrac{hc}{E}$. Thus, careful observation leads to correct answers.