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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= 2piħn, 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) |
| Answer» <html><body><p></p>Solution :(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` <br/> or `p = pi ħ//l`<br/> Hence`E_(n) = (p^(2))/(2 m) = (pi^(2) n^(2) ħ)/(2 m l)`. <br/> `oint p d x = 2 l p` because we have to consider the <a href="https://interviewquestions.tuteehub.com/tag/integral-516833" style="font-weight:bold;" target="_blank" title="Click to know more about INTEGRAL">INTEGRAL</a> form `- (1)/(2)` to `(1)/(2)` and then back to `-(1)/(2)`. <br/> (b) Here, `oint p d x = 2 pi r p = 2 pi nħ` <br/> or `p = (n ħ)/(r )` <br/> Hence`E_(n) = (n^(2)ħ^(2))/(2 m r^(2))` <br/> (c ) By energy <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a>`(p^(2))/(2 m) + (1)/(2) alpha x^(2) = E` <br/> so`p = sqrt(2m E - m alpha x^(2))` <br/> Then `oint p d x = oint sqrt(2m E - m alpha x^(2) dx)` <br/> `= 2sqrt(m alpha) int_(-(sqrt( 2E))/(alpha))^(sqrt(2E)/(alpha)) sqrt((2 E)/(alpha) - x^(2)) dx` <br/> The integral is`int_(-a)^(a)sqrt(a^(2) - x^(2)) dx = a^(2) int_(-x//2)^(x//2) cos^(2) theta d theta` <br/> `= (a^(2))/(2) int_(-x//2)^(x//2)(1 + cos 2 theta) d theta = a^(2)(pi)/(2)`. <br/> Thus `oint p d x = pi sqrt(m a). (2 E)/(alpha) = E.2 pi sqrt((m)/(alpha)) = 2 pi n ħ` <br/> Hence `E_(n) = n ħsqrt((alpha)/(m))`. <br/> ( b) It is required to find the energy levels of the <a href="https://interviewquestions.tuteehub.com/tag/circular-916697" style="font-weight:bold;" target="_blank" title="Click to know more about CIRCULAR">CIRCULAR</a> orbit for the rotential <br/> `U( r) = -(alpha)/(r )` <br/> In a circular orbit, the particle only has <a href="https://interviewquestions.tuteehub.com/tag/tangible-663303" style="font-weight:bold;" target="_blank" title="Click to know more about TANGIBLE">TANGIBLE</a> velocity and the qunatization condition reads `ointp d x = m v. 2 pi r = 2 pi nħ` <br/> so `m v r = M = n ħ` <br/> The energy of the particle is <br/> `E = (n^(2) ħ^(2))/(2 m r^(2)) - (alpha)/(r )` <br/> Equilibrium requires that the energy as a function of `r` be minimum. Thus <br/> `(n^(2) ħ^(2))/(mr^(3))=(alpha)/(r^(2)) or r=(n^(2) ħ^(2))/(m alpha)` <br/> Hence `E_(n)= -(malpha^(2))/(2n^(2) ħ^(2))`</body></html> | |