I came across this question in text book and was not sure how to solve it. When momentum is conserved then $$ m_{H_2} \Delta v_{H_2} = -m_{Photon} \Delta v_{Photon} $$

But then photon has a zero rest mass which eventually leads to $\Delta v_{H_2}=0$, doesn't it. And when momentum is conserved does it not mean that the energy of the molecule stays the same? But then that is the basis for vibration and rotation spectroscopy.

enter image description here


1 Answer 1


The problem is that a the momentum of a photon is not its rest mass multiplied by its velocity. Rather it is $p = \frac{h}{\lambda} = \frac{E}{c}$ where h is the Planck constant and $\lambda$ the wavelength, E the energy of the photon, and c the speed of light.

You can find a decent explanation for this for example here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.