Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is :

Dual Nature of Radiation PYQ — JEE Main 2020 · 3 Sep · Shift 2 · Q61 with Solution

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Correct answer: A

For light of wavelength $\lambda$ , the energy of one photon is

$E = \frac{hc}{\lambda}$

So, shorter wavelength means higher energy per photon.

Number of photons emitted per second is

$N = \frac{P}{E} = \frac{P}{hc/\lambda} = \frac{P\lambda}{hc}$

Thus,

$N \propto \lambda$

when power is same.

So,

$\frac{N_X}{N_V} = \frac{\lambda_X}{\lambda_V} = \frac{1}{500}$

For radiation traveling in space,

$\text{number density} = \frac{\text{number crossing per second}}{\text{volume crossing per second}}$

Since both travel with speed $c$ , number density is proportional to photon flux, hence proportional to photon emission rate for equal power beams.

Therefore,

$\frac{n_X}{n_V} = \frac{N_X}{N_V} = \frac{1}{500}$

Therefore, the correct answer is:

$\boxed{\frac{1}{500}}$