hw2sol

# hw2sol - ignores the atom recoil 6(b atom momentum p = Mv...

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6 (a) Energy conservation: E i = E f + hf + p 2 2 M , (1) where p is the momentum of the recoiling atom. Momentum conservation: 0 = p γ - p (2) where p γ is the momentum of the emitted photon γ . But p γ = hf c (3) Hence, (1) - (2) give: E i = E f + hf + h 2 f 2 2 Mc 2 , (4) i.e. the exact formula for f is: f 2 + 2 Mc 2 h f - 2 Mc 2 h 2 ( E i - E f ) = 0 . (5) Thus solving for f we get the answer: f = - Mc 2 h + Mc 2 h ± 1 + 2 Mc 2 ( E i - E f ) ² 1 / 2 . (6) Note : Clearly, only one sign can be used in solving the quadratic equation (5) for f . Why? Now the atom mass and rest anergy Mc 2 is much larger that any electron energies involved: 2 Mc 2 ( E i - E f ) << 1 Hence expanding the square root by Taylor expansion: [1 + x ] 1 / 2 = 1 + 1 2 x - 1 8 x 2 + O ( x 3 ) , x << 1 we get from (6): f = ( E i - E f ) h - 1 2 ( E i - E f ) 2 hMc 2 + O ( ( E i - E f ) 3 h ( Mc 2 ) 2 ) . (7) The Frst term is precisely the Bohr formula, and we see that the corrections to it go like 1 /Mc 2 . If one takes the M → ∞ limit one of course gets just the Bohr formula that

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Unformatted text preview: ignores the atom recoil. 6 (b) atom momentum p = Mv . After photon emission and recoil one has from (2): v = p M = p γ M = hf Mc , (8) 1 where the exact answer for f is given by (6), or, in the form expanded in inverse powers of mc 2 , by (7). If in (8) we use only the Frst term from (7), i.e. the Bohr answer, we get the approximate answer you are supposed to derive, i.e. v = h Mc ( E i-E f ) h = Rh Mc 1 n 2 i-1 n 2 f ! . (9) The second and higher terms in (7) when inserted in (8) will give the corrections to (9), which we see go like O (1 /M 2 c 3 ). 2...
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hw2sol - ignores the atom recoil 6(b atom momentum p = Mv...

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