By separating the angular and radial components of the Wavefunction
\begin{align} \Psi(x, y, z) \rightarrow R(r) Y(\phi, \theta) \end{align}the radial wavefunction for a Hydrogenic atom takes the form
\begin{align} -\frac{1}{2 \mu} \frac{d^2 u(r)}{dr^2} + V_{eff}u(r) &= E u(r) \\ u(r) &= rR(r) \\ V_{eff}(r) &= -\frac{1}{r} + \frac{\hbar^2 l (l + 1)}{2 \mu r^2} \end{align}where $\mu$ is the reduced mass.
Find the analytic expression for $R_{nl}(r)$ at Explain Everything and make Python Functions for n = 1, l = 0, n = 2, l = 0, 1 and n = 3, l = 0, 1, 2.
Note that on Expliain Everything the functions are given as a function of $\rho$. The relation between $\rho$ and $r$ is: $\rho = \frac{2 Z}{n a}r$ where $a = 4\pi \epsilon_0 \frac{ \hbar^2 }{\mu e^2}$
def R10(r):
# n = 1, l = 0
Bonus points to those who can make a General R(n, l, r).
def R(n, l, r):
I was not able to do this so please tell me if you come up with a solution.
Now let's study the radial probablilty density $P(r)$, the total wavefunction is $\Psi_{nlm_l}$. To study the radial probability density we take the probability density $|\Psi_{nlm_l}|$ and integrate the non-radial parts ($\theta, \phi$).
\begin{align} P(r) = \int_{surface} |\Psi_{nlm_l}(\theta, \phi, r)| d\tau = \int_0^\pi \int_0^{2 \pi} R_{nl}(r)^2 |Y_{lm_l}(\theta, \phi)| r^2\sin (\theta) d\theta d \phi \end{align}we then find $P(r) = R_{nl}(r)^2 r^2$
Plot the Quantity $P(r) = R_{nl}(r)^2 r^2 $ for various values of Z.
What happens when Z increases, why do you think this happens?
The total wavefunciton $Y_{n l m_l} = R_{nl}(r) Y_{lm_l}(\theta, \phi)$ where $R_{nl}$ is the solution to the Radial wavefunciton and
Make a function that returns the probability density of finding the particle within a given volume $\theta \in (\theta_0, \theta_1)$ $r \in (r_0, r_1)$$\phi \in (\phi_0, \phi_1)$
def two_pz_probability(theta_0, theta_1, phi_0, phi_1, r_0, r_1):
In [ ]: