Question

A gas absords a photon of $355nm$ and emits at two wavelengths. If one of the emission is at $680nm$, then the other would be at:

• A. $1035nm$
• B. $352nm$
• C. $743nm$
• D. $518nm$

Answer 1

The correct option is C $743nm$

By using conservation of energy
$h\upsilon =h{\upsilon }_1+h{\upsilon }_2$

Also we have $\upsilon =c\lambda$ where $\upsilon$ is frequency , $\lambda$ is wavelength, h is plancks constant and c is speed of light

thus we get the equation

$\frac{1}{\lambda }=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$
If we put $\lambda =355nmand{\lambda }_1=680nm$
$\frac{1}{355}=\frac{1}{680}+\frac{1}{{\lambda }_2}$

${\lambda }_2=743nm$
conclusion: Hence the answer is option c =$743nm$