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# hw-9 - m 1 2 = ~ 2 r k m 1 2(c This dierence is given by E...

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HW # 9 Solutions 1. The following was graded as correct. Using the Heisenberg uncertainty relation p x h 4 π we set ∆ x = 1Angstrom = 10 1 0 m, which is the side of the box. This gives us p h 4 π x p 5 . 27 × 10 2 5 kg m s 1 Then, using the expression for the kinetic energy of the particle, we find that the minimum energy carried by the particle is E = p 2 2 m = (5 . 27 × 10 2 5 kg m s 1 ) 2 (2 × 10 3 0 kg) = 13 . 89 × 10 4 0 J E m i n = (13 . 89 × 10 2 0 ) (1 . 602 × 10 1 9 ) eV = 0 . 867eV 2(a) The equation of motion for the classical harmonic oscillator is F = - k x = m d 2 x dt 2 which yields the general solution x ( t ) = A sin(2 π f t + B ) where A and B are constants, and f = 1 2 π r k m is the frequency. In the above, k is the spring’s constant, and m is the mass. Thus, the classical frequency is given by f = 1 2 π r k m . 2(b) The ground state is given for n = 0. With the angular frequency, ω = 2 π f and ~ = h/ (2 π ) the ground state energy is E = ~ ω 1 2 = ~ 2 π 1 2 π r k m 1 2 = ~ 2

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Unformatted text preview: m 1 2 = ~ 2 r k m 1 2(c) This dierence is given by E n =2-E n =0 = ~ (2 + 1 / 2)-~ (1 / 2) = 2 ~ = 2 ~ 2 f = 2 ~ 2 1 2 r k m = 2 ~ r k m 2(d) In this case, the energy of the emitted photon is E n =1-E n =0 = ~ (1 + 1 / 2)-~ (1 / 2) = ~ = ~ 2 f = ~ 2 1 2 r k m = ~ r k m But we also know that E = ~ c , so that we get that = c p k/m = c r m k 2(e) We may use the mass of the electron here, m = 9 . 11 10 31 kg. Then, using c = 3 . 10 8 m s 1 and k = 10eV / (Angstrom) 2 = 160 . 2J m 2 , we nd = c r m k = 22 . 62 10 9 m = 22 . 62 nm which lies outside the visible spectrum; it lies in the ultra-violet part of the electromagnetic spectrum. 2...
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hw-9 - m 1 2 = ~ 2 r k m 1 2(c This dierence is given by E...

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