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Q.
A gas absorbs a photon of $355$ nm and emits at two wavelengths. If one of the emissions is at $680$ nm, then find the wavelength of the other in nm.
Structure of Atom
Solution:
Energy of absorbed photon = Sum of the energies of emitted photon
$\frac{hc}{\lambda} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}$
or $\frac{1}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}$
$ \frac{1}{355 \times10^{-9}} = \frac{1}{680 \times10^{-9}} +\frac{1}{\lambda}$
$\frac{1}{\lambda_{2}} = \frac{1}{355 \times10^{-9}} - \frac{1}{680 \times10^{-9}} = 1.346\times10^{6}$
or $\lambda_{2} =1 1.346 \times10^{6} = 743 \times10^{-9}$ m
$=743$ nm