If a source of 4 kW power emits 10^{20} phot | vedclass

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(B) The power $P$ of the source is given by $P = n 	imes E = n 	imes (hc / \lambda)$,where $n$ is the number of photons per second,$h$ is Planck's c

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$X$-rays

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(B) The power $P$ of the source is given by $P = n 	imes E = n 	imes (hc / \lambda)$,where $n$ is the number of photons per second,$h$ is Planck's constant,and $c$ is the speed of light.Given $P = 4 	imes 10^3 \ W$ and $n = 10^{20} \ s^{-1}$.Rearranging for wavelength: $\lambda = (n 	imes h 	imes c) / P$.Using $hc \approx 1.987 	imes 10^{-25} \ J \cdot m$:$\lambda = (10^{20} 	imes 1.987 	imes 10^{-25}) / (4 	imes 10^3) \approx 0.5 	imes 10^{-8} \ m = 50 \ \mathring{A}$.Since the wavelength $\lambda = 50 \ \mathring{A}$ falls in the range of $0.1 \ \mathring{A}$ to $100 \ \mathring{A}$,this radiation belongs to the $X$-ray region.

If a source of $4 \ kW$ power emits $10^{20}$ photons/second,then this radiation belongs to which part of the electromagnetic spectrum?

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