The Real Hydrogen Atom
Age 16 to 18
Challenge Level
This problem is really pushing well into
university concepts, but the solution given here is well worth a
read by keen school students who wish to further their
understanding of differential equations.
Task 0: The lowest energy
set of equation corresponds to $ n = 0$. This implies that $l =0$
and $m_l = 0$ also.
Thus:
a) $\frac{1}{R} \frac{d}{dr}\left[r^2 \frac{dR}{dr}\right] -\frac{8
\pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$
b) $\frac{sin\theta}{P} \frac{d}{d\theta} \left[sin\theta
\frac{dP}{d\theta}\right] = 0$
c) $\frac{1}{F} \frac{d^2F}{d\phi^2} = 0$
Task 1: If $R_1$ is a
solution, then for $aR_1$:
$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right]
-\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$
Since $a$ is a constant, it can be removed outside of the
differential:
$\frac{a}{aR_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right]
-\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$
$\frac{1}{R_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right]
-\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$
Therefore, since $R_1$ is a solution, so is $aR_1$ for any non-zero
constant $a$.
If $P_1$ is a solution, then for $bP_1$:
$\frac{sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta
\frac{d(bP_1)}{d\theta}\right] = 0$
Since $b$ is a constant, it can be removed outside of the
differential:
$\frac{b\ sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta
\frac{dP_1}{d\theta}\right] = 0$
$\frac{sin\theta}{P_1} \frac{d}{d\theta} \left[sin\theta
\frac{dP_1}{d\theta}\right] = 0$
Therefore, since $P_1$ is a solution, so is $bP_1$ for any non-zero
constant $b$.
If $F_1$ is a solution, then for $cF_1$:
$\frac{1}{cF_1} \frac{d^2(cF_1)}{d\phi^2} = 0$
Since $c$ is a constant, it can be removed outside of the
differential:
$\frac{c}{cF_1} \frac{d^2F_1}{d\phi^2} = 0$
$\frac{1}{F_1} \frac{d^2F_1}{d\phi^2} = 0$
Therefore, since $F_1$ is a solution, so is $cF_1$ for any non-zero
constant $c$.
Task 2: If R is a constant,
then $\frac{dR}{dr} = 0$
$\therefore \frac{1}{R} \frac{d}{dr}\left[0\right] -\frac{8 \pi^2
\mu}{h^2}[Er^2 +\alpha r] = l(l+1)$
$ -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$
Since $r > 0$, and $l> 0$, this equation is not satisfied by
a constant value of $R$.
If P is a constant, then $\frac{dP}{d\theta} = m_l^2$
$\therefore \frac{sin\theta}{P} \frac{d}{d\theta} \left[0\right] +
l(l+1)sin^2\theta= m_l^2$
$l(l+1)sin^2\theta = m_l^2$
The equation is not satisfied since there is still a dependence on
$\theta$, which means that the wavefunction would only be permitted
for certain values of $\theta$ Therefore a constant value of $P$ is
not permissible.
If F is a constant, then:
$m_l^2 = 0$
$m_l = 0$
Task 3:
a) Let $P = sin\theta$
$\therefore \frac{dP}{d\theta} = cos\theta$
$\rightarrow \frac{sin\theta}{sin\theta} \frac{d}{d\theta}
\left[sin\theta cos\theta \right] + l(l+1)sin^2 \theta=
m_l^2$
$cos(2\theta) + l(l+1)sin^2 \theta = m_l^2$
Expanding $cos (2\theta)$ as $cos^2\theta - sin^2\theta$, and
simplifying gives:
$cos^2\theta + (l^2 +l -1)sin^2\theta = m_l^2$
To eliminate the trigonometric terms, we wish to use the identity
$cos^2\theta + sin^2\theta = 1$
$\therefore l^2 + l -2 = 0$
$(l+2)(l-1) = 1$
$\mathbf{ l =1}$ since $l > 0$
$\mathbf{m_l = \pm 1}$
b) Now let $P = cos\theta$
$\therefore \frac{dP}{d\theta} = -sin\theta$
$\frac{sin\theta}{cos\theta}\frac{d}{d\theta}\left[-sin^2\theta\right]
+ l(l+1)sin^2\theta = m_l^2$
Differentiating, simplifying and collecting terms yields:
$(l^2 + l -2) sin^2\theta = m_l^2$
For this to be valid, $\mathbf{m_l^2 = 0}$, which means that either
$sin^2 \theta$ or $(l^2 + l -2)$ must be equal to zero.
Clearly, for the wavefunction to exist:
$l^2+ 1 -2 = 0$
$\mathbf{l=1}$ since $l > 0$
c) Substituting in $P = tan\theta$ yields an equation still
containing trigonometry. There is no way for this to be removed,
and so the original trial solution is invalid.
If you haven't tried any other trigonometric functions yet, why not
try $cos^2\theta$, $sin(2\theta)$ and $sin^2\theta$...?
Task 4:
$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right]
-\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$
Given that $a_0 = \frac{\pi e^2 \mu}{h^2 \epsilon_0}$, $\alpha =
\frac{e^2}{4\pi \epsilon_0}$, $E = -\frac{\mu
e^4}{8n^2h^2\epsilon_0^2}$:
$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right]
-\frac{8 \pi^2 \mu}{h^2}[-\frac{\mu e^4 r^2}{8n^2h^2\epsilon_0^2}
+\frac{e^2r}{4\pi \epsilon_0}] = l(l+1)$
$\frac{1}{R}\frac{d}{dr} \left[r^2\frac{dR}{dr}\right] +
(\frac{a_0r}{n})^2 -2a_0r = l(l+1)$