logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Atomic Structure

A photon of 300nm is absorbed by a gas and then re-emits two photons. One re-emitted photon has wavelength 496nm. Calculate energy of other photon re-emitted out.

$(a)\;3.62\times10^{-19}J\qquad(b)\;1.623\times 10^{-19}J\qquad(c)\;2.625\times 10^{-19}J\qquad(d)\;4.8\times 10^{-19}J$

1 Answer

$E_{photon-absorbed}=\large\frac{6.625\times10^{-34}\times3.0\times 10^8}{300\times10{-9}}$
$E_{re-emitted}=\large\frac{6.625\times10^{-34}\times3.0\times 10^8}{496\times10^{-9}}$
$=4.0\times10^{-19}J$
$E_{absorbed} = E_{photon}+E_{photon re-emitted}$
$E_{11} = 6.625\times 10^{-19}-4.0\times 10^{-19}$
$=2.625\times10^{-19}$
Hence the answer is (c)
answered Feb 4, 2014 by sharmaaparna1
 

Related questions

Download clay6 mobile appDownload clay6 mobile app
...
X