Question

A blue-green photon (λ = 488 nm ) is absorbed by a free hydrogen atom, initially at rest. What is the recoil speed of the hydrogen atom after absorbing the photon?

Answer 1

Answer:

The recoil speed is $2.207\times 10^{4} m/s$

Solution:

Wavelength of a blue-green photon, $\lambda_{BG} = 488 nm = 488\times 10^{- 9} m$

Now, the energy associated with the blue-green photon:

$E_{BG} = \frac{hc}{\lambda_{BG}}$

where

h = Planck's constant

C = speed of light ion vacuum

$E_{BG} = \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{488\times 10^{- 9}}$

$E_{BG} = 4.07\times 10^{- 19} J$

Also, we know that the recoil speed can be calculated by the KInetic energy which is equal to the Energy of the blue-green photon:

$KE_{H} =\frac{1}{2}m_{p}v_{H}$

where

$v_{H}$ = velocity of Hydrogen atom

$m_{p} = 1.67\times 10^{- 27} kg$ = mass of H-atom

Now,

$KE_{H} =\frac{1}{2}m_{p}(v_{H})^{2}$

$4.07\times 10^{- 19} =\frac{1}{2}\times 1.67\times 10^{- 27}\times (v_{H})^{2}$

$v_{H} = \sqrt(4.87\times 10^{8}) = 2.207\times 10^{4} m/s$