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11.Dual Nature of Radiation and matter
medium
If a source of power $4\,kW$ produces $10^{20}$ photons/second, the radiation belongs to a part of the spectrum called
A
$\gamma$ -rays
B
$X$ -rays
C
Ultraviolet rays
D
Microwaves
Solution
No. of photons emitting per second from a source of power $P$ is
$n = \left( {5 \times {{10}^{24}}} \right)P\lambda $
wavelength emitting $\lambda=\frac{\mathrm{n}}{\left(5 \times 10^{24}\right) \mathrm{P}}\left[\mathrm{or} \lambda=\frac{\mathrm{nhc}}{\mathrm{P}}\right]$
$ \Rightarrow \lambda = \frac{{{{10}^{20}}}}{{5 \times {{10}^{24}} \times 4 \times {{10}^3}}} = 0.5 \times {10^{ – 8}}{\rm{m}} = 50\mathop {\rm{A}}\limits^o $
And this wavelength comes in $X$ ray region.
Standard 12
Physics
