- #1
Chen
- 977
- 1
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:
[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. I guessed that 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
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. I guessed that 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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