Solving Atom Velocity in Non-relativistic Limit

[Chen]

Hello,
I have an atom of mass M which is on an energy level of E1. After emitting a photon, it comes down to a lower level E0. The question is - what's the velocity of the atom after the photon emission? And it says to simplify the expression using the "non-relativistic limit". I'm not sure what the last phrase means, can anyone please explain?

Here's my solution of the problem:

$$\Delta E = E_1 - E_0$$

From conservation of momentum we have:

$$M v + \frac{h \nu}{c} = 0$$

$$\nu = -\frac{M c v}{h}$$

And from conservation of energy:

$$\Delta E = h\nu + \frac{M v^2}{2}$$

Substituting $$\nu$$ we get:

$$M v^2 - 2 M c v - 2\Delta E = 0$$

$$v_{1,2} = \frac{2 M c +/- \sqrt{4 M^2 c^2 + 8 M \Delta E}}{2 M}$$

Since $$v < c$$ we must choose:

$$v = c - \sqrt{c^2 + \frac{2 \Delta E}{M}}$$

And now I need to simplify this expression. I guessed that the limit should be:

$$\frac{\Delta E}{M c^2}$$ -> 0

But I'm really not sure. Does anyone have any idea?


Thanks,
Chen

 

[Physics Monkey]

Yes, the nonrelativistic limit assumes that the final velocity of the atom is much smaller than c. Equivalently, the change in energy is small compared to the rest mass energy. You should expand that last expression in a power series.

 

[Chen]

Great, thanks. Although ideologically, I like to avoid power series whenever possible:

$$v = c - \sqrt{c^2 + \frac{2 \Delta E}{M}} = \frac{c^2 - c^2 - \frac{2 \Delta E}{M}}{c + \sqrt{c^2 + \frac{2 \Delta E}{M}}} = \frac{- \frac{2 \Delta E}{M c}}{1 + \sqrt{1 + \frac{2 \Delta E}{M c^2}}} = - \frac{\Delta E}{M c}$$

Chen

 

[Physics Monkey]

Don't like power series? But power series are your friends! Why don't you want to hang out with your friends? Haha.

 

[Chen]

They're ok, but it would be such an overkill to expand this expression into a series... :)