Question

A near ultraviolet photon of 300 nm is absorbed by a gas and then remitted as two photons. One photon is red with wavelength of 760 nm. What would be the wave number of the second photon?

• A. $2\times {10}^6{m}^{-1}$
• B. $2\times {10}^4{m}^{-1}$
• C. $2\times {10}^5{m}^{-1}$
• D. $2\times {10}^{11}{m}^{-1}$

Answer 1

The correct option is A $2\times {10}^6{m}^{-1}$

$\text{Energy absorbed = Sum of energy of two quanta}$
Since, E = $\frac{hc}{\lambda }$
$\frac{hc}{300\times {10}^{-9}}=\frac{hc}{760\times {10}^{-9}}+\frac{hc}{\lambda }$
$\frac{1}{300\times {10}^{-9}}=\frac{1}{760\times {10}^{-9}}+\frac{1}{\lambda }$
On solving, we get $\lambda =4.95\times {10}^{-7}$ m
Now, wave number $\stackrel{-}{v}$ = $\frac{1}{\lambda }$
$\stackrel{-}{v}=\frac{1}{\lambda }=2\times {10}^6{m}^{-1}$