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

The recoil speed is
`2.207* 10^(4) m/s`

Solution:

Wavelength of a blue-green photon,
`\lambda_(BG) = 488 nm = 488* 10^(- 9) m`

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

`E_(BG) = (hc)/(\lambda_(BG))`

where

h = Planck's constant

C = speed of light ion vacuum

`E_(BG) = (6.626* 10^(- 34)* 3* 10^(8))/(488* 10^(- 9))`

`E_(BG) = 4.07* 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) =(1)/(2)m_(p)v_(H)`

where

`v_(H)` = velocity of Hydrogen atom

`m_(p) = 1.67* 10^(- 27) kg` = mass of H-atom

Now,

`KE_(H) =(1)/(2)m_(p)(v_(H))^(2)`

`4.07* 10^(- 19) =(1)/(2)* 1.67* 10^(- 27)* (v_(H))^(2)`

`v_(H) = \sqrt(4.87* 10^(8)) = 2.207* 10^(4) m/s`