# Collisional excitation minimum velocity

1. Apr 27, 2013

### torchflame

1. The problem statement, all variables and given/known data
A sodium atom emits a photon with wavelength 818 nm shortly after being struck by an electron. What minimum speed did the electron have before the collision?

2. Relevant equations
$E_{particle}\geq\Delta E_{atom}$

3. The attempt at a solution
For the minimum energy, $E_{particle}=\Delta E_{atom}$, and because the photon was emitted due to the collision:
$\frac{m_ev^2}{2}=\frac{hc}{\lambda}$
$m_ev^2=\frac{2hc}{\lambda}$
$v=\sqrt{\frac{2hc}{m_e \lambda}}$
$v=\sqrt{\frac{2 \times 1239.842 \mathrm{eV} \times \mathrm{nm}}{9.11 \times 10^{-31} \times 818 \mathrm{nm} \times \mathrm{kg}}}$

When I evaluate this, I get $.73 \times 10^6 \frac{\mathrm m}{\mathrm s}$, which Mastering Physics tells me is wrong. Is there something obvious I'm missing here?

2. Apr 27, 2013

### Staff: Mentor

The sodium atom will move after the collision, this needs some energy. I don't know which precision is required here.

3. Apr 27, 2013

### torchflame

Okay, so if I look at the collision from the perspective of conservation of momentum, and I say $M$ is the mass of the sodium atom, I get that $m_ev_0+0M=m_ev_{post e}+Mv_{post Na}$.
If I take the minimum energy, $v_{poste}=0$. That implies that the velocity of the sodium after the collision: $v_{pS}=\frac{m_ev_0}{M}$. Then I think I have to assume that $E_{imparted}=E_{electron}-E_{Na}$?
If I assume that, I get that $E_{Na}=\frac{m_e^2v_0^2}{2M}$, which implies that $\frac{hc}{\lambda}=v_0^2\frac{2Mm_e-2m_e^2}{4M}$ which would imply that $v_0^2=\frac{hc}{\lambda}\frac{4M}{2Mm_e-2m_e^2}$, which gives me $v_0=730192 \frac{\mathrm{m}}{\mathrm{s}}$, which Mastering Physics also says is wrong.
Help?

4. Apr 28, 2013

### Staff: Mentor

Hmm.... I agree with your value of ~730km/s, and as you calculated, those corrections are small. I really don't know which precision Mastering Physics wants (maybe it is just a matter of the correct input format?). For the minimal energy, both the nucleus and the electron should have the same velocity afterwards, and the photon will carry some momentum. Those are really tiny modifications of the result, however.

5. Apr 30, 2013

### torchflame

Well, apparently the answer was $v_0=1.16 \times 10^6 \frac{\mathrm{m}}{\mathrm{s}}$, which I have absolutely no idea how they got.
Anyone have any ideas?

Last edited: Apr 30, 2013