(a) $\psi$ depends only on $r$, so §8.3a gives $\hat{\mathbf L}\psi = 0$: zero, in every component and in magnitude.

(b) Both are standard integrals with the hint’s formula. The kinetic one needs the radial Laplacian; both give the stated results.

(c) Minimizing

$$\langle E\rangle = \frac{\hbar^2}{2m_ea^2} - \frac{e^2}{4\pi\epsilon_0a}$$

gives $\mathrm d\langle E\rangle/\mathrm da = -\hbar^2/m_ea^3 + e^2/4\pi\epsilon_0a^2 = 0$, so

$$a = \frac{4\pi\epsilon_0\hbar^2}{m_ee^2} = a_0 \qquad\text{and}\qquad \langle E\rangle_{\min} = -\frac{e^2}{8\pi\epsilon_0a_0} = -E_R = -13.6\ \mathrm{eV}$$

This is the whole ground state, obtained without solving anything. The trial function happens to be exact here — but the method works even when it is not, and it always gives an upper bound on the true ground-state energy.

(a) Setting the derivative to zero gives $r_c$ directly. At that radius the particle sits at the bottom of the effective potential with $p_r = 0$ and no way to move radially — a circular orbit, and the lowest energy Eq. (9.2) allows for that $L$.

(b) Setting $p_r = 0$ in Eq. (9.2) gives a quadratic in $r$ whose roots are the two stated turning points — perihelion and aphelion. When $a = r_c$ the discriminant vanishes and the two coincide: the circular orbit of part (a).

Compare with quantum mechanics. $r_c = l(l+1)a_0$ is Eq. (9.18)‘s minimum, and problem 3(e) shows the quantum state approaches it as $l \to \infty$. But the quantum ground state at $l = 0$ has no classical counterpart at all — a classical electron with zero angular momentum falls straight in.

Parts (a)–(d) are the hint’s integral applied four times.

(e) is the point of the problem. As $l$ grows:

• the most probable radius $(l+1)^2a_0 \to l^2a_0$;
• the mean radius $\tfrac12(2l+3)(l+1)a_0 \to l^2a_0$;
• and problem 2’s classical $r_c = l(l+1)a_0 \to l^2a_0$.

All three converge. Meanwhile part (d) gives $\Delta r/\langle r\rangle \to 0$: the state stops being spread out and becomes a sharp shell at a definite radius.

That is the correspondence principle made concrete. A state of large $l$ with no radial nodes is a quantum object that behaves like a classical circular orbit — sharp radius, definite energy, and the classical radius to boot. §8.1’s cone picture said the same thing about direction; this says it about position.

(a) $\int_0^\infty e^{-2r/a_0}4\pi r^2\mathrm dr = 4\pi\cdot\dfrac{2}{(2/a_0)^3} = \pi a_0^3$, so $N_1 = 1/\sqrt{\pi a_0^3}$.

(b) The integral is a sum of two of problem 1’s integrals with $\alpha = 3/2a_0$; requiring it to vanish fixes $\lambda = -1/2a_0$, which is exactly Table 9.1’s 2s state.

What this shows is not arithmetic. Eigenfunctions of a Hermitian operator with different eigenvalues are automatically orthogonal — §7.1 proved it. So this integral had to vanish, and the calculation is really determining $\lambda$: the 2s state is forced into its form by orthogonality to 1s, not by anything about the potential.

(a) $n = 1$: only $l = 0$, giving 1 state $= 1^2$. $n = 2$: $l = 0$ gives 1 and $l = 1$ gives 3, total $4 = 2^2$.

(b) Going from $k$ to $k+1$ adds exactly one new value of $l$, namely $l = k$, contributing $2k+1$ states. So

$$g_{k+1} = g_k + (2k+1) = k^2 + 2k + 1 = (k+1)^2$$

The induction works because $(k+1)^2 - k^2 = 2k+1$ is precisely the size of the multiplet being added — an identity about squares doing the work of a physical argument.

Doubling for spin gives $2n^2$, which is the shell capacity 2, 8, 18, 32 that chapter 11 builds the periodic table from.

Substituting $n_r = n - l - 1$ into Eq. (9.25) collapses it to

$$\langle r\rangle = \frac{a_0}{2}\left[3n^2 - l(l+1)\right]$$

which §9.3 verified is algebraically identical to the book’s form.

state⇅	$n$⇅	$l$⇅	$\langle r\rangle / a_0$⇅
1s	1	0	$1.5$
2s	2	0	$6$
2p	2	1	$5$ⓘ
3s	3	0	$13.5$
3p	3	1	$12.5$
3d	3	2	$10.5$ⓘ

Mean radii from the collapsed formula. Click a cell for detail.

The pattern is $3n^2$ dominating, so size grows quadratically with $n$ while $l$ makes a smaller correction downward.

Chapter 7’s problem 8 gave $2\langle T\rangle = \langle r\,\mathrm dV/\mathrm dr\rangle$, and for $V = -k/r$, $r\,\mathrm dV/\mathrm dr = +k/r = -V$. So $2\langle T\rangle = -\langle V\rangle$. Combined with $E_n = \langle T\rangle + \langle V\rangle$:

$$\langle T\rangle = -E_n = +\frac{E_R}{n^2},\qquad \langle V\rangle = 2E_n = -\frac{2E_R}{n^2}$$

For the ground state: $\langle T\rangle = +13.6$ eV and $\langle V\rangle = -27.2$ eV.

The consequence is worth carrying: $E = -\langle T\rangle$, so a hydrogen atom that loses energy has its electron speed up. That is the same algebra that gives self-gravitating systems a negative heat capacity, and ch07’s problem 8 drew the same conclusion for satellites and stars.

The angular integrals give $4\pi\hbar\sin(pr/\hbar)/pr$, leaving a radial integral that is elementary, and the stated amplitude follows.

The distribution has no sharp cutoff. $|\tilde\psi|^2$ falls off as $p^{-8}$, so arbitrarily large momenta occur with small probability — the electron is localized to $\sim a_0$, so its momentum must be spread. That is the uncertainty principle appearing as a fact about the shape of a function rather than as an inequality.

And the peak is not the mean again: $1/\sqrt3 = 0.577$ against $8/3\pi = 0.849$. The momentum distribution is skewed for exactly the reason the radial one is — §9.3’s caution applies verbatim.

$2s \to 1s$. Both states are spherically symmetric, so the integrand is (even) × (odd) × (even) = odd, and integrating an odd function over all space gives zero. This is §9.4’s parity argument on its simplest possible case.

$2p \to 1s$. Now $\psi_{2p} \propto z$ (for $m_l = 0$), and the operator contributes another $z$, so the integrand goes as $z^2$ — even, and the integral survives.

The two cases differ only in parity, which is the entire content of the selection rule. And the consequence is the metastability of 2s that §9.4 computed: 0.14 s against 1.6 ns.

With $l = 1$, $s = \tfrac12$, the bracket gives $+1$ for $j = \tfrac32$ and $-2$ for $j = \tfrac12$ — the 1 : −2 ratio in the stated answers. Multiplying by $\langle 1/r^3\rangle$ for the 2p state gives the coefficients.

The difference is

$$\frac{1}{96} - \left(-\frac{2}{96}\right) = \frac{3}{96} = \frac{1}{32}\,\alpha^4m_ec^2 = 45.3\ \mu\mathrm{eV}$$

and dividing that into the 10.2 eV transition energy shifts the 121.5 nm line by $5.4\times10^{-4}$ nm — the book’s answer, confirmed.

(a) $\mu = \dfrac{m_em_e}{m_e+m_e} = \dfrac{m_e}{2}$ — exactly half, the most extreme case possible.

(b) $E_n = \dfrac{\mu}{m_e}\left(-\dfrac{E_R}{n^2}\right) = -\dfrac{E_R}{2n^2}$, so the ground state is $-6.8$ eV rather than $-13.6$ eV.

(c) Since $a_0' = (m_e/\mu)a_0 = 2a_0$, the mean radius is $\tfrac32 \times 2a_0 = \mathbf{3a_0}$.

Half the binding, twice the size. And note what positronium is not: there is no heavy centre, so “the electron orbits the positron” is exactly as wrong as the reverse. The reduced mass handles that automatically.

(a) $\mu = \dfrac{207m_e\,m_p}{207m_e + m_p} = 186\,m_e$ — not 207, because the proton is only about 9 times heavier than the muon and recoils appreciably.

(b) $E_n = -186\,E_R/n^2$, so the ground state is $-2530$ eV — an X-ray transition rather than an ultraviolet one.

(c) $\langle r\rangle = \tfrac32a_0/186 = 0.0081\,a_0 = 2.8\times10^{-13}$ m.

A muonic atom is 186 times smaller and 186 times more tightly bound. The muon orbits so close that the proton is no longer a good point charge — which is what problem 13 is about, and it is why muonic hydrogen is used to measure the proton’s radius.

Using the corrected $\tfrac43(R/a_0)^3$:

system	effective $a_0$	$P(r < R_{\text{proton}})$
ordinary hydrogen	$5.29\times10^{-11}$ m	$7\times10^{-14}$
muonic hydrogen	$2.85\times10^{-13}$ m	$5\times10^{-7}$

A factor of about $6\times10^6$, because the probability goes as $a_0^{-3}$ and the muonic radius is 186 times smaller: $186^3 \approx 6\times10^6$.

That is why muonic atoms measure nuclear size. The energy shift from the proton not being a point charge is proportional to how much of the wave function sits inside it. For an electron that is a part in $10^{13}$ — unmeasurable. For a muon it is a part in $10^6$, which is a large effect in modern spectroscopy, and muonic hydrogen is how the proton radius is measured today.

Energies and eigenfunctions. Everywhere $a_0$ appears, replace it by $a_0/Z$; everywhere $E_R$ appears, replace it by $Z^2E_R$:

$$E_n = -\frac{Z^2E_R}{n^2},\qquad u_{0,0}(r) = \frac{2}{\sqrt{a_0/Z}}\left(\frac{Zr}{a_0}\right)e^{-Zr/a_0}$$

and similarly for 2s and 2p.

Ionization energies. He⁺ has $Z = 2$: $4 \times 13.6 = \mathbf{54.4}$ eV. Li²⁺ has $Z = 3$: $9 \times 13.6 = \mathbf{122.5}$ eV.

Why relativity matters more. The electron’s speed scales as $Z\alpha c$, so the relativistic correction — which goes as $(v/c)^4$ — scales as $Z^4\alpha^4$. Going from hydrogen to Li²⁺ multiplies it by $3^4 = 81$.

For a heavy element like uranium ($Z = 92$), $Z\alpha \approx 0.67$: the inner electrons are genuinely relativistic, and no non-relativistic treatment works at all. This is why gold is yellow and mercury is liquid — both are relativistic effects on the outer electrons of heavy atoms.

The physics is ch04’s problem 8. The decay is far faster than the electron can respond, so its wave function is unchanged at that instant — but the Hamiltonian is not. The probability of finding it in a particular new eigenstate is the modulus squared of the overlap:

$$P = \left|\int\psi^{\mathrm{He}^+}_{1s}(\mathbf r)^*\,\psi^{\mathrm H}_{1s}(\mathbf r)\,\mathrm d^3\mathbf r\right|^2$$

Using $\psi^{\mathrm H}_{1s} \propto e^{-r/a_0}$ and $\psi^{\mathrm{He}^+}_{1s} \propto e^{-2r/a_0}$ (problem 14), the exponentials combine to $e^{-3r/a_0}$ and the constants give the stated prefactor.

The integral is $2/(3/a_0)^3 = \tfrac{2}{27}a_0^3$, so

$$P = 128\left(\frac{2}{27}\right)^2 = \frac{512}{729} = \mathbf{0.702}$$

So 70% of the time the electron lands in the new ground state, and 30% of the time it does not — ending up excited, or ionized outright. Nothing pushed it; the Hamiltonian simply changed underneath a wave function that had no time to adjust.