If a source of power 4 , kW produces 10^{20} | vedclass

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(B) The power $P$ of a source emitting $n$ photons per second with wavelength $\lambda$ is given by $P = \frac{nhc}{\lambda}$.Rearranging for waveleng

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$X$-rays

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(B) The power $P$ of a source emitting $n$ photons per second with wavelength $\lambda$ is given by $P = \frac{nhc}{\lambda}$.Rearranging for wavelength: $\lambda = \frac{nhc}{P}$.Given: $P = 4 \, kW = 4 	imes 10^3 \, W$, $n = 10^{20} \, 	ext{photons/s}$, $h = 6.63 	imes 10^{-34} \, J \cdot s$, and $c = 3 	imes 10^8 \, m/s$.Substituting the values: $\lambda = \frac{10^{20} 	imes 6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{4 	imes 10^3}$.$\lambda = \frac{19.89 	imes 10^{-6}}{4 	imes 10^3} = 4.97 	imes 10^{-9} \, m \approx 50 \, \mathring{A}$.Since the wavelength is approximately $50 \, \mathring{A}$, this radiation falls in the $X$-ray region of the electromagnetic spectrum.

If a source of power $4 \, kW$ produces $10^{20}$ photons/second, the radiation belongs to a part of the spectrum called

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