# Two sources of light emit X-rays of wavelength $1\; nm$ and visible light of wavelength $500\; nm$, respectively. Both the sources emit light of the same power $200\; W$. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is : Option: 1 Option: 2 Option: 3 Option: 4

The energy of each photon is given by

$E=h \nu =\frac{h c}{\lambda} \ \ (in \ Joule)$

And

n ( the number of photons emitted per second) is given as

$n=\frac{Power \ of \ source \ (W \ or \ \frac{J}{sec} )}{Energy \ of \ each \ photon (J)}=\frac{P}{E} \ (sec^{-1})$

So $P=nE$

$Here, P=100 W, \lambda_{1}=1 n m, \lambda_{2}=500 \mathrm{nm}\\$

Let $n_{1}, n_{2}=$ no. of photons X -rays and visible light emitted from the two sources

$\therefore P=n_{1} \frac{h c}{\lambda_{1}}=n_{2} \frac{h c}{\lambda_{2}} \\ \Rightarrow \frac{n_{1}}{\lambda_{1}}=\frac{n_{2}}{\lambda_{2}} \\ \Rightarrow \frac{n_{1}}{n_{2}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{1}{500}$

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