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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 \frac{1}{500}
Option: 2 250
Option: 3 \frac{1}{250}
Option: 4 500

Answers (1)

best_answer

 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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avinash.dongre

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