Photon Energy that can Produce an Electronic Transition

[MathIsFun]

1. The problem statement, all variables, and given/known data
After thinking over your beloved but erratic instructor’s lectures on the Bohr model of the atom, you come to the sad but inevitable conclusion that he has, yet again, lied to you. You have been taught that to make an electronic transition between two states differing by an energy $\Delta E$, the atom must absorb a photon with this energy. Demonstrate that this cannot be true by conserving momentum and energy to calculate the energy of a photon that can produce an electronic transition $\Delta E$ in an atom (in the atom’s rest frame), expressing the result in terms of the ratio of the energy of the photon to the energy of the transition, and derive the Taylor expansion for this quantity to linear terms in $\Delta E$ (to sufficient accuracy so that a deviation from $E= \Delta E$ is noted). You can ignore relativistic effects, e.g., the energy of the photon and $\Delta E$ are much, much less than the rest mass of the atom. What is this ratio for the $\textrm{H}\alpha$ transition?

Homework Equations

$$E(n,Z)=E_{0} \frac{Z^2}{n^2}, E_{0}=-\frac{k_{e}^2 e^4 m_{e}}{2\hbar^2}$$ $$v(n,Z)=v_{0} \frac{Z}{n}, v_{0}=\frac{k_{e} e^2}{\hbar}$$ $$E_{n,Z}=-\frac{m_{e}v_{n,Z}^2}{2}$$ $$E_{0}=-\frac{m_{e} v^2_{0}}{2}$$ $$\Delta p_{tot}=0$$ $$\Delta E_{tot}=0$$ $$E_{\gamma}=h f_{\gamma}=\frac{hc}{\lambda_{\gamma}}$$ $$p_{\gamma}=\frac{E_{\gamma}}{c}$$

The Attempt at a Solution

I started with conservation of momentum, assuming that the photon would transfer all of its energy to the electron (I'm not sure if this assumption is correct).
$$\frac{E_{\gamma}}{c}+m_{e}v_{ei}=0+m_{e}v_{ef}$$ $$E_{\gamma}=c\Delta p_{e}$$
Since $\Delta p_{e}=Zm_{e}v_{0}\left(\frac{1}{n_{f}}-\frac{1}{n_{i}}\right)$ and $\Delta E_{e}=Z^2 E_{0}\left(\frac{1}{n_{f}^2}-\frac{1}{n_{i}^2}\right)=Z^2 E_{0}\left(\frac{1}{n_{f}}-\frac{1}{n_{i}}\right)\left(\frac{1}{n_{f}}+\frac{1}{n_{i}}\right)$, $\Delta p_{e}=\frac{\Delta E_{e} m_{e}v_{0}}{Z E_{0} \left(\frac{1}{n_{i}}+\frac{1}{n_{f}}\right)}=-\frac{2\Delta E_{e}}{Z v_{0}\left(\frac{1}{n_{i}}+\frac{1}{n_{f}}\right)}$.
Then $E_{\gamma}=-\frac{2c \Delta E_{e}}{Z v_{0} \left(\frac{1}{n_{i}}+\frac{1}{n_{f}}\right)}$.
So $\frac{E_{\gamma}}{\Delta E_{e}}=-\frac{2c}{Z v_{0} \left(\frac{1}{n_{i}}+\frac{1}{n_{f}}\right)}=-\frac{2c\hbar}{Z k_{e} e^2\left(\frac{1}{n_{i}}+\frac{1}{n_{f}}\right)}$. $$\frac{E_{\gamma}}{\Delta E}=-\frac{2c\hbar}{Z k_{e} e^2 \left(\frac{1}{n_{i}}+\frac{1}{n_{f}}\right)}$$
First of all, I'm not sure if this is correct (just plugging in $n_{i}=1$ and $n_{f}=2$ gives a wild result). Second, I would have to do a Taylor series for this equation in terms of $\Delta E$, but I'm not sure how I would do that since it only depends on a linear term of $\Delta E$ (which supports my original thought that this is incorrect).
Calculating this value for the $\mathrm{H}\alpha$ transition is only a matter of plugging in $Z=1$, $n_{i}=3$, and $n_{f}=2$.

I think that there is something incorrect in my initial assumption when I used conservation of momentum. How should I have started this problem? Where did I make a mistake?

Thanks

 

[mfb]

The photon transfers all its momentum to the atom. Excitations are properties of atoms, not of electrons.

$E_{\gamma}=c\Delta p_{e}$

While technically correct (apart from the "e" subscript), evaluating the momentum change that way is quite messy and probably not what you want to do.

$\Delta p_{e}=Zm_{e}v_{0}\left(\frac{1}{n_{f}}-\frac{1}{n_{i}}\right)$

That equation doesn't make sense.

For now, assume that the energy change in the rest frame of the atom needs an energy of $\Delta E$. Ignore Bohr's model, the problem is much more general.

 

[MathIsFun]

So I have $E_{\gamma}=m_{a} c \Delta v_{a}$ from conservation of momentum, but I don't know how I would relate this with $\Delta E$.

 

[mfb]

This is conservation of momentum. You still have conservation of energy you can include.
Note that m won't stay exactly the same, but the difference is small enough to neglect it here.