Question

A photon of 300 nm is absorbed by a gas, which then re-emits two photons. One re-emitted photon has a wavelength of 400 nm. Calculate the energy of the other photon re-emitted out.

• A. $16.56\times {10}^{-18}J$
• B. $16.56\times {10}^{-12}J$
• C. $16.56\times {10}^{-20}J$
• D. $16.56\times {10}^{-15}J$

Answer 1

The correct option is C

$16.56\times {10}^{-20}J$

By Planck's Theory,

$E=\frac{hc}{\lambda }$

Now, $E_{Total}=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{300\times {10}^{-9}m}$

= $6.626\times {10}^{-19}J$

$E_1=\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{400\times {10}^{-9}m}$

= $\frac{3}{4}\times 6.626\times {10}^{-19}J$

$E_2=(E_{Total}-E_1)$

= $6.626\times {10}^{-19}\times (1-\frac{3}{4})=\frac{1}{4}\times 6.626\times {10}^{-19}J$

= $1.656\times {10}^{-19}J$