Q63 / 75 · JEE Main 2026 (08 April Shift 2) · Physics
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).
- AMicrowave
- BInfrared
- CVisible
- Correct answerDUltraviolet
Solution
Power of the source, $P = 15 \text{ kW} = 15 \times 10^3 \text{ J/s}$
Number of photons emitted per second, $n = 2.5 \times 10^{22} \text{ s}^{-1}$
Energy of one photon, $E = \dfrac{P}{n} = \dfrac{15 \times 10^3}{2.5 \times 10^{22}} = 6 \times 10^{-19} \text{ J}$
Using $E = \dfrac{hc}{\lambda}$, we get $\lambda = \dfrac{hc}{E}$
$\lambda = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{6 \times 10^{-19}} = 3.3 \times 10^{-7} \text{ m} = 330 \text{ nm}$
The wavelength $330 \text{ nm}$ lies in the ultraviolet region of the electromagnetic spectrum.
Answer: Ultraviolet
Number of photons emitted per second, $n = 2.5 \times 10^{22} \text{ s}^{-1}$
Energy of one photon, $E = \dfrac{P}{n} = \dfrac{15 \times 10^3}{2.5 \times 10^{22}} = 6 \times 10^{-19} \text{ J}$
Using $E = \dfrac{hc}{\lambda}$, we get $\lambda = \dfrac{hc}{E}$
$\lambda = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{6 \times 10^{-19}} = 3.3 \times 10^{-7} \text{ m} = 330 \text{ nm}$
The wavelength $330 \text{ nm}$ lies in the ultraviolet region of the electromagnetic spectrum.
Answer: Ultraviolet
Hence, the correct answer is option: 'D'.