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(C) The power $P$ of the source is $15 	ext{ kW} = 15 	imes 10^3 	ext{ W}$.The number of photons produced per second $n = 10^{16} 	ext{ s}^{-1}$.T

Vedclass · Accepted Answer

Gamma rays

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(C) The power $P$ of the source is $15 	ext{ kW} = 15 	imes 10^3 	ext{ W}$.The number of photons produced per second $n = 10^{16} 	ext{ s}^{-1}$.The energy of one photon $E$ is given by $E = \frac{P}{n} = \frac{15 	imes 10^3}{10^{16}} = 15 	imes 10^{-13} 	ext{ J}$.Using the relation $E = h
u$,we find the frequency $
u$:$
u = \frac{E}{h} = \frac{15 	imes 10^{-13}}{6 	imes 10^{-34}} = 2.5 	imes 10^{21} 	ext{ Hz}$.Electromagnetic radiation with a frequency of $2.5 	imes 10^{21} 	ext{ Hz}$ falls in the range of Gamma rays (which typically have frequencies greater than $10^{19} 	ext{ Hz}$).

If a source of electromagnetic radiation having power $15 \text{ kW}$ produces $10^{16}$ photons per second,the radiation belongs to which part of the spectrum? (Take Planck constant $h = 6 \times 10^{-34} \text{ Js}$)

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