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

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Question

No. of photons emitting per second from a source of power $P$ is

$n = \left( {5 	imes {{10}^{24}}} ight)P\lambda $

wavelength emitting $\lamb

Vedclass · Accepted Answer

$X$ -rays

Vedclass · Answer

No. of photons emitting per second from a source of power $P$ is

$n = \left( {5 	imes {{10}^{24}}} ight)P\lambda $

wavelength emitting $\lambda=\frac{\mathrm{n}}{\left(5 	imes 10^{24}ight) \mathrm{P}}\left[\mathrm{or} \lambda=\frac{\mathrm{nhc}}{\mathrm{P}}ight]$

$ \Rightarrow \lambda  = \frac{{{{10}^{20}}}}{{5 	imes {{10}^{24}} 	imes 4 	imes {{10}^3}}} = 0.5 	imes {10^{ - 8}}{m{m}} = 50\mathop {m{A}}\limits^o $

And this wavelength comes in $X$ ray region.

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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