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 :

Solution

Physics sir · 3 steps

Same power means same energy per second, so the source with the cheaper photons sends out more of them.

1. Power in photons. Each photon carries energy E = hc/λ. If a source emits N photons per second, its power is P = Nhc/λ, so N = Pλ/hc.
2. Number density. Both kinds of photon travel at the same speed c. The number per unit volume is therefore proportional to the number emitted per second, and the same P, h and c cancel in the ratio.

   n_X/n_V = λ_X/λ_V = (1 nm)/(500 nm) = 1/500

3. So. Each X-ray photon carries 500 times the energy of a visible photon, so only 1/500 as many are needed for the same 200 W. The answer is (a).

The idea to keep. At fixed power, the number of photons is proportional to λ. Shorter wavelength means more energy per photon, so fewer photons.

Answer from the NTA final answer key. Solution by PrepWiser.