OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 14A / ii / Solution

For fixed

l

, the radial equation is

- \frac{ℏ ^{2}}{2 μ r ^{2}} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + \frac{ℏ ^{2} l ( l + 1 )}{2 μ r ^{2}} R - \frac{e ^{2}}{4 π ϵ _{0} r} R = ER .

Substitute

R (r) = C r^{l} e^{- r / b} .

Cancellation of the coefficient of

1/ r

requires

b = (l + 1) a, a = \frac{4 π ϵ _{0} ℏ ^{2}}{μ e ^{2}},

the reduced-mass Bohr radius. The remaining constant term gives

E_{l} = - \frac{ℏ ^{2}}{2 μ a ^{2} ( l + 1 ) ^{2}} = - \frac{e ^{2}}{8 π ϵ _{0} a ( l + 1 ) ^{2}} .

Each such no-radial-node state is a circular Coulomb bound state with principal quantum number

n = l + 1

.

Solved by gpt-5.6-sol high.

 Ancestors (11)