# Classical atomic models

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# A gas absorbs a near  photon of wavelength and then emitted as $2$ photons. One photon is red which has a wavelength of $760$ Calculate the wavelength of second photon.

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Solution

## ${\lambda }_{1}$ (U.V. light) $=300\mathrm{nm}=300×{10}^{-9}\mathrm{m};{\lambda }_{2}$(red light) $=760\mathrm{nm}=760×{10}^{-9}\mathrm{m}$.For U.V. light, $E=\frac{hc}{\lambda }=\frac{hc}{300×{10}^{-9}\mathrm{m}}$                           …(i)For red light, ${E}_{1}=\frac{hc}{{\lambda }_{1}}=\frac{hc}{760×{10}^{-9}\mathrm{m}}$                           …(ii)Energy associated with second photon, ${E}_{2}$              Now $E={E}_{1}+{E}_{2}$.$\begin{array}{l}\frac{hc}{300×{10}^{-9}\mathrm{m}}=\frac{hc}{760×{10}^{-9}\mathrm{m}}+\frac{hc}{{\lambda }_{2}}\\ \frac{hc}{300×{10}^{-9}\mathrm{m}}=hc\left[\frac{1}{760×{10}^{-9}\mathrm{m}}+\frac{1}{{\lambda }_{2}}\right]\\ \frac{1}{{\lambda }_{2}}=\frac{1}{300×{10}^{-9}\mathrm{m}}-\frac{1}{760×{10}^{-9}\mathrm{m}}\\ =\frac{\left(760×{10}^{-9}\mathrm{m}\right)-\left(300×{10}^{-9}\mathrm{m}\right)}{300×{10}^{-9}\mathrm{m}×760×{10}^{-9}\mathrm{m}}\\ =\frac{460×{10}^{-9}\mathrm{m}}{228,000×{10}^{-18}{\mathrm{m}}^{2}}\\ {\lambda }_{2}=\frac{228000×{10}^{-18}{\mathrm{m}}^{2}}{460×{10}^{-9}\mathrm{m}}\\ =4.96×{10}^{-7}\mathrm{m}\\ =4.96×{10}^{-7}\mathrm{m}×\frac{1\mathrm{nm}}{{10}^{-9}\mathrm{m}}\end{array}$.

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