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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.

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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.

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