Question 6.P.4: An electron is trapped inside an infinite spherical well V(r...

An electron is trapped inside an infinite spherical well V(r)=\begin{cases} 0, & r<a, \\ +\infty, & r>a.\end {cases}.

(a) Using the radial Schrödinger equation, determine the bound eigenenergies and the corresponding normalized radial wave functions for the case where the orbital angular momentum of the electron is zero (i.e., l = 0).

(b) Show that the lowest energy state for l = 7 lies above the second lowest energy state for l = 0.

(c) Calculate the probability of finding the electron in a sphere of radius a/2, and then in a spherical shell of thickness a/2 situated between r = a and r = 3a/2.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

(a) Since V(r) = 0 in the region r ≤ a, the radial Schrödinger equation (6.57)

-\frac{\hbar ^{2} }{2M} \frac{d^{2}U_{nl}(r) }{dr^{2} } +V_{eff} (r)U_{nl}(r)=E_{n} U_{nl}(r),           (6,57)

becomes

-\frac{\hbar ^{2} }{2m}\left[\frac{d^{2}U_{nl}(r)}{dr^{2}}-\frac{l(l+1)}{r^{2}} U_{nl}(r) \right] =EU_{nl}(r),               (6.241)

where U_{nl}(r)=rR_{nl}(r). For the case where l = 0, this equation reduces to

\frac{d^{2}U_{n0}(r)}{dr^{2}} =-k^{2}_{n} U_{n0}(r),                     (6.242)

where k^{2}_{n}=2mE_{n} /\hbar ^{2}. The general solution to this differential equation is given by

U_{n0}(r)=A\cos (k_{n}r )+B\sin (k_{n}r )                      (6.243)

or

R_{n0}(r)=\frac{1}{r} (A\cos (k_{n}r )+\sin (k_{n}r )).                    (6.244)

Since R_{n0}(r) is finite at the origin or U_{n0}(0)= 0, the coefficient A must be zero. In addition, since the potential is infinite at r = a (rigid wall), the radial function R_{n0}(a) must vanish:

R_{n0}(a)=B\frac{\sin k_{n}a}{a} =0;                    (6.245)

hence ka=n\pi ,n=1, 2, 3, …. This relation leads to

E_{n} =\frac{\hbar ^{2}\pi ^{2} }{2ma^{2}} n^{2}.                    (6.246)

The normalization of the radial wave function R(r), ∫^{a}_{0} \left|R_{n0}(r)\right| ^{2} r^{2}dr=1, leads to

1=\left|B\right| ^{2} ∫^{a}_{0} \frac{1}{r^{2}} \sin ^{2} (k_{n}r )r^{2}dr=\frac{\left|B\right| ^{2}}{k_{n}} ∫^{k_{n}a}_{0} \sin ^{2}\rho d\rho =\frac{\left|B\right| ^{2}}{k_{n}}\left(\frac{\rho }{2} -\frac{\sin 2\rho }{4} \right) \mid ^{\rho =k_{n}a}_{\rho =0}

 

=\frac{1}{2} \left|B\right| ^{2}a;                                              (6.247)

hence B=\sqrt{2/a} . The normalized radial wave function is thus given by

R_{n0}(r)=\sqrt{\frac{2}{a} } \frac{1}{r} \sin \left(\sqrt{\frac {2mE_{n} }{\hbar ^{2}} }r \right).                   (6.248)

(b) For l = 7 we have

E_{1} (l=7)>V_{eff}(l=7)=\frac{56\hbar ^{2}}{ma^{2}} =\frac{28\hbar ^{2}}{ma^{2}}.                             (6.249)

The second lowest state for l = 0 is given by the 3s state; its energy is

E_{2} (l=0)=\frac{2\pi ^{2}\hbar ^{2} }{ma^{2}} ,                       (6.250)

since n = 2. We see that

E_{1} (l=7)>E_{2} (l=0).                                      (6.251)

(c) Since the probability of finding the electron in the sphere of radius a is equal to 1, the probability of finding it in a sphere of radius a/2 is equal to 1/2.

As for the probability of finding the electron in the spherical shell between r = a and r = 3a/2, it is equal to zero, since the electron cannot tunnel through the infinite potential from r < a to r > a.

Related Answered Questions