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

$E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}Where\phantom{\rule{0ex}{0ex}}Eisenergy\phantom{\rule{0ex}{0ex}}hisPlanck\text{'}scons\mathrm{tan}t\phantom{\rule{0ex}{0ex}}cisspeedoflight\phantom{\rule{0ex}{0ex}}\lambda iswavelength$

According to law of conservation of energy,

$\frac{hc}{355}=\frac{hc}{680}+\frac{hc}{{\lambda}_{2}}\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}\frac{680-355}{355x680}=\frac{1}{{\lambda}_{2}}\phantom{\rule{0ex}{0ex}}{\lambda}_{2}=742.77nm$

= 743 nm

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