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A gas absorbs a photon of 355 nm and emits at two wavelengths. if one of the emissionsis at 680 nm, the other is at

nikhil , 10 Years ago
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Sunil Kumar FP

Last Activity: 10 Years ago

energy corresponding to wavelength 355nm=hc/l
where h= planks constant
c=speed of light
l=wavelength
therefore according to the problem,
hc/355*10^-9=hc/680^10^-9 +hc/l*10^-9
1/355=1/680+1/l
l=742nm
Thus the wavelength of the light for other emmission is 742nm

Tanishka Thorat

Last Activity: 5 Years ago

As we know c=v (lambda).. Symbol of not available  for lamba
By plank's theory, 
E=hv=>E=hc/lambda
Now, using the law of conservation of energy  : absorbed energy =emitted energy 
hc/lambda=hc/lambda1+hc /lambda 2 =>1/lamdba=1/lambda1+1/lambda 2
=>1/355=1/680+1/lambda2
=>1/lmabda 2=1/133-1/680
=>lambda2=743nm

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