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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:

[tex]\Delta E = E_1 - E_0[/tex]

From conservation of momentum we have:

[tex]M v + \frac{h \nu}{c} = 0[/tex]

[tex]\nu = -\frac{M c v}{h}[/tex]

And from conservation of energy:

[tex]\Delta E = h\nu + \frac{M v^2}{2}[/tex]

Substituting [tex]\nu[/tex] we get:

[tex]M v^2 - 2 M c v - 2\Delta E = 0[/tex]

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

Since [tex]v < c[/tex] we must choose:

[tex]v = c - \sqrt{c^2 + \frac{2 \Delta E}{M}}[/tex]

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

[tex]\frac{\Delta E}{M c^2}[/tex] -> 0

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

Thanks,

Chen

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# Homework Help: Non-relativistic limit

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