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(B) The power of the source is given by $P = n \cdot E_{photon} = n \cdot \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.Fr

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$X -$ rays

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(B) The power of the source is given by $P = n \cdot E_{photon} = n \cdot \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.From this,the wavelength $\lambda$ is given by $\lambda = \frac{nhc}{P}$.Substituting the given values: $n = 10^{20}$,$h = 6.625 	imes 10^{-34} \, J \cdot s$,$c = 3 	imes 10^8 \, m/s$,and $P = 4 	imes 10^3 \, W$.$\lambda = \frac{10^{20} 	imes 6.625 	imes 10^{-34} 	imes 3 	imes 10^8}{4 	imes 10^3} \, m$.$\lambda = \frac{19.875 	imes 10^{-6}}{4 	imes 10^3} \, m = 4.96875 	imes 10^{-9} \, m \approx 5 	imes 10^{-9} \, m$.This wavelength range ($10^{-8} \, m$ to $10^{-11} \, m$) corresponds to $X -$ rays.

The power of a source is $4 \, kW$. If it produces $10^{20}$ photons in $1 \, s$,then the radiation emitted from this source belongs to which part of the electromagnetic spectrum?

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