InterviewSolution
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InterviewSolution
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According to the Bohr-Sommerfeld postulate the periodic motion of a particle in a potential field must satisfy the following qyantization rule: `ointp dq= 2pi ħn`, where `q` and `p` are generalized coordinate and momenum of the particle, `n` are integers. Making use of this rule, find the permitted values of energy for a particle of mass `m` moving (a) In a uniimensional rectangular potential well of width `l` (b) along a circule of radius `r`, (c ) in a unidimentional potential field `U=alphax^(2)//2`, where `alpha` is a positive constant: (d) along a round orbit in a central field, where the potential enargy of the particle is equal to `U= -alpha//r`(`alpha` is a positve constant) |
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Answer» (a) if we measure energy from the bottom of the well, then `V(x) = 0` inside the walls. Then the quantization condition reads `oint p d x = 2 l p = 2 pi nħ` or `p = pi ħ//l` Hence `E_(n) = (p^(2))/(2 m) = (pi^(2) n^(2) ħ)/(2 m l)`. `oint p d x = 2 l p` because we have to consider the integral form `- (1)/(2)` to `(1)/(2)` and then back to `-(1)/(2)`. (b) Here, `oint p d x = 2 pi r p = 2 pi n ħ` or `p = (n ħ)/(r )` Hence `E_(n) = (n^(2) ħ^(2))/(2 m r^(2))` (c ) By energy conservation `(p^(2))/(2 m) + (1)/(2) alpha x^(2) = E` so `p = sqrt(2m E - m alpha x^(2))` Then `oint p d x = oint sqrt(2m E - m alpha x^(2) dx)` `= 2sqrt(m alpha) int_(-(sqrt( 2E))/(alpha))^(sqrt(2E)/(alpha)) sqrt((2 E)/(alpha) - x^(2)) dx` The integral is `int_(-a)^(a)sqrt(a^(2) - x^(2)) dx = a^(2) int_(-x//2)^(x//2) cos^(2) theta d theta` `= (a^(2))/(2) int_(-x//2)^(x//2)(1 + cos 2 theta) d theta = a^(2)(pi)/(2)`. Thus `oint p d x = pi sqrt(m a). (2 E)/(alpha) = E.2 pi sqrt((m)/(alpha)) = 2 pi n ħ` Hence `E_(n) = n ħsqrt((alpha)/(m))`. ( b) It is required to find the energy levels of the circular orbit for the rotential `U( r) = -(alpha)/(r )` In a circular orbit, the particle only has tangible velocity and the qunatization condition reads `oint p d x = m v. 2 pi r = 2 pi n ħ` so `m v r = M = n ħ` The energy of the particle is `E = (n^(2) ħ^(2))/(2 m r^(2)) - (alpha)/(r )` Equilibrium requires that the energy as a function of `r` be minimum. Thus `(n^(2) ħ^(2))/(mr^(3))=(alpha)/(r^(2)) or r=(n^(2) ħ^(2))/(m alpha)` Hence `E_(n)= -(malpha^(2))/(2n^(2) ħ^(2))` |
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