Question

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

• A. 1035 nm
• B. 325 nm
• C. 743 nm
• D. 518 nm

Answer 1

The correct option is C

743 nm

Energy of abosrbed photon = Sum of the energies of emitted photon

$\frac{hc}{\lambda }$ = $\frac{hc}{{\lambda }_1}$ + $\frac{hc}{{\lambda }_2}$ or $\frac{1}{\lambda }$ = $\frac{1}{{\lambda }_1}$ + $\frac{1}{{\lambda }_2}$

$\frac{1}{355\times {10}^{-9}}$ = $\frac{1}{680\times {10}^{-9}}$ + $\frac{1}{{\lambda }_2}$

$\frac{1}{{\lambda }_2}$ = $\frac{1}{355\times {10}^{-9}}$ - $\frac{1}{680\times {10}^{-9}}$ = 1.346 $\times$ ${10}^5$

${\lambda }_2$ = $\frac{1}{1.346}$ ${10}^5$ = 743 $\times$ ${10}^{-9}$ m = 743nm