A source of cm wave of 400 W produces \( 10^{20} \) photons per second. Identify the part of the electromagnetic spectrum to which these radiations belongs to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → ultraviolet rays
The power of the source is the energy carried per second by its photons, so the energy of one photon is the power divided by the number of photons per second. Its wavelength then follows from \( E = hc/\lambda \).
\( \displaystyle E = \frac{P}{N} = \frac{400\,\mathrm{W}}{10^{20}\,\mathrm{s^{-1}}} = 4\times10^{-18}\,\mathrm{J} \)
\( \displaystyle \lambda = \frac{hc}{E} = \frac{(6.63\times10^{-34})(3\times10^{8})}{4\times10^{-18}} \approx 5\times10^{-8}\,\mathrm{m} = 50\,\mathrm{nm} \)
A wavelength of about 50 nm lies in the ultraviolet region (roughly 400 nm down to about 1 nm). (In electron volts, \( E = 4\times10^{-18}/1.6\times10^{-19} = 25\,\mathrm{eV} \).)