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:

We start by recalling the basic relation for the energy of a single photon. The energy carried by one photon of wavelength $$\lambda$$ is given by Planck’s formula

$$E_{\text{photon}} = h\,\nu = \dfrac{h\,c}{\lambda},$$

where $$h$$ is Planck’s constant and $$c$$ is the speed of light in vacuum.

Next, we connect this to the power of a source. Power $$P$$ is the total energy emitted per unit time. If a source emits $$N$$ photons each second, then the power can be written as

$$P = N \, E_{\text{photon}}.$$

Substituting $$E_{\text{photon}} = \dfrac{h\,c}{\lambda}$$, we obtain

$$P = N \left( \dfrac{h\,c}{\lambda} \right).$$

Solving for the photon emission rate (number of photons emitted each second), we get

$$N = \dfrac{P\,\lambda}{h\,c}.$$

Observe that, for a fixed power $$P$$, the number of photons emitted per second is directly proportional to the wavelength $$\lambda$$.

Both the X-ray source and the visible-light source have the same power $$P = 200\,\text{W}$$. Hence, the ratio of their photon emission rates is simply the ratio of their wavelengths:

$$\dfrac{N_{\text{X-ray}}}{N_{\text{visible}}} = \dfrac{\lambda_{\text{X-ray}}}{\lambda_{\text{visible}}}.$$

The wavelengths are given as

$$\lambda_{\text{X-ray}} = 1\,\text{nm}, \qquad \lambda_{\text{visible}} = 500\,\text{nm}.$$

Substituting these values, we find

$$\dfrac{N_{\text{X-ray}}}{N_{\text{visible}}} = \dfrac{1\,\text{nm}}{500\,\text{nm}} = \dfrac{1}{500}.$$

Therefore, the number density (or equivalently, the photon emission rate per unit time) of X-ray photons is smaller by a factor of $$500$$ compared with that of the visible-light photons.

Hence, the correct answer is Option A.