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

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(B) The power $P$ of the source is given by $P = n \cdot E_{photon} = n \cdot h 
u$,where $n$ is the number of photons per second,$h$ is Planck's con

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$X$-rays

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(B) The power $P$ of the source is given by $P = n \cdot E_{photon} = n \cdot h 
u$,where $n$ is the number of photons per second,$h$ is Planck's constant,and $
u$ is the frequency.Given $P = 4 \ kW = 4 	imes 10^3 \ W$ and $n = 10^{20} \ photons/s$.Rearranging for frequency: $
u = \frac{P}{n \cdot h}$.Substituting the values: $
u = \frac{4 	imes 10^3}{10^{20} 	imes 6.63 	imes 10^{-34}} \approx 6.03 	imes 10^{16} \ Hz$.The frequency range $6 	imes 10^{16} \ Hz$ falls within 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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