A monochromatic source of light operating at 15 kW emits $$2.5 \times 10^{22}$$ photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to __________. (Take $$h = 6.6 \times 10^{-34}$$ J.s and $$c = 3 \times 10^8$$ m/s).

Solution :

Power emitted by source :

$$P = 15\text{ kW}$$

$$= 1.5 \times 10^4\text{ W}$$

Number of photons emitted per second :

$$n = 2.5 \times 10^{22}\text{ photons/s}$$

Energy of one photon :

$$E = \frac{P}{n}$$

$$=\frac{1.5 \times 10^4}{2.5 \times 10^{22}}$$

$$= 0.6 \times 10^{-18}$$

$$= 6 \times 10^{-19}\text{ J}$$

Using photon energy relation :

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

Therefore,

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

$$=\frac{(6.6 \times 10^{-34})(3 \times 10^8)}{6 \times 10^{-19}}$$

$$=\frac{19.8 \times 10^{-26}}{6 \times 10^{-19}}$$

$$= 3.3 \times 10^{-7}\text{ m}$$

$$= 330\text{ nm}$$

Wavelength 330nm lies in ultraviolet region.

Final Answer :

Ultraviolet region.