Recoil velocity of atom due to electron orbital change

Question

If electron jumps from n =1 to n = 2 in hydrogen atom then how will we find recoil velocity of atom ( I am just beginner and had never crossed such question so can't understand where to start)

Answer 1

There are three things to conserve during an electronic transition: energy, momentum, and angular momentum. As always, these things are related to some extent. Let's forget about the angular momentum because it isn't relevant to this question.

The energy of a photon is given by,$$E_\gamma=h\nu.$$This energy is related to the momentum of a photon through the mass-energy equivalence,$$E^2=m^2c^4+p^2c^2.$$ Remembering that the mass of a photon is zero, we have,$$P_\gamma=\frac{E_\gamma}{c}.$$

The energy of an electronic transition between two states in a hydrogen atom must equal the energy of the photon being absorbed, which conserves the energy upon either absorption or emission. Thus, we have,$$E_\gamma=\frac{m_ee^4}{8h^2\epsilon_0^2}\left (\frac{1}{n_1^2}-\frac{1}{n_2^2}\right ),$$where we have used the solution to the energy states of a hydrogen atom coming from using a Coulomb potential to describe the potential energy of the system.

Then, we impose the conservation of momentum condition that $P_\gamma=P_{atom,f}-P_{atom,i}$. When I write $P_{atom}$ we have to note that this is given by the sum of momenta of the nucleus and electron.

Let's define the recoil velocity as, $v_{recoil}=v_f-v_i$

I don't wanna deal with the special relativistic correction to the momentum, so I'll leave that to you if you wanna figure it out.

Now, we can write,$$P_{recoil}=\frac{E_\gamma}{c},$$which can be written in terms of the recoil velocity according to,$$v_{recoil}=\frac{E_\gamma}{(m_e+m_p)c}.$$In the last expression, $m_p$ is the mass of a proton (i.e. the mass of the nucleus).

The expression would get uglier if you used $P=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$ instead of $P=mv$.

Using the above expression, the recoil velocity ends up being,$$v_{recoil}=\frac{10.2\cdot10^{-5}MeV}{938.272\frac{MeV}{c}}=3.26\ \frac{m}{s}.$$

So that's a reasonable velocity I think.

It's totally possible I made a mistake somewhere because I've never thought about this question before, but I think that is all correct. It's a little bit weird to think about whether the momentum is being transferred to the electron or the nucleus, but it seems like it must be transferred to both somehow?

One way to check the work (which I don't wanna do) is to try to get the same expression by only balancing energies and using $\frac{1}{2}mv^2$.

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This question was also asked on chem.SE so I reposted my answer here. I'm sure someone will correct any mistake I made.

Answer 2

From momentum conservation, $p_{\text{photon}} = p_{\text{hydrogen}} \implies \color{grey}{\dfrac hλ} = mv$

From energy conservation,

$ΔE = E_{\text{photon}} + KE_{\text{hydrogen}}$

KE gained by the hydrogen atom is very small and can be ignored.

$ΔE = hυ = \dfrac{\color{grey}{h} \ c}{\color{grey}{λ}} = mvc$

So $v = \dfrac {ΔE}{mc} = \frac{13.6(1 - \frac 1{2^2})1.6 \ 10^{-19}}{1.67 \ 10^{-27} 3\ 10^8} = 3.264 \ \text{m/s}$