A professor once asked us in class: for the Hydrogen atom, what is the probability of finding the electron inside the nucleus?

The radius of a proton (the nucleus of a Hydrogen atom) is $r_0 \approx 10^{-15}\textrm{m}$. The answer to the question will be hard to find if one tries to compute it analytically by integrating the probability density function $|\psi({\bf x})|^2$ over the volume of the proton: $$ P = \int_{\|{\bf x}\| \leq r_0} |\psi({\bf x})|^2 d{\bf x} \label{post_ec0bf1e2547df878846dd841b8d657b2_eq_prob_r0} $$ where $\|{\bf x}\| \leq r_0$ refers to the fact that the integral should be computed over a sphere of radius $r_0$ centered at the origin (where the center of the proton is assumed to be). Instead, as my professor told us, a much better idea is to compute $P$ "as Fermi would". By that, he meant the problem can be easily solved using a very good approximation: assume that $|\psi({\bf x})|$ is constant over the extension of the nucleus. This is indeed an excellent approximation which turns the computation of the desired probability into trivial task: $$ \boxed{ \displaystyle P \approx |\psi({\bf 0})|^2 \int_{\|{\bf x}\| \leq r_0} d{\bf x} = \frac{4\pi r_0^3}{3}|\psi({\bf 0})|^2 } \label{post_ec0bf1e2547df878846dd841b8d657b2_eq_prob_approx1} $$

For any state for which the wavefunction $\psi({\bf x})$ is known and varies little within the extension of the nucleus (this is typically true), one can use equation \eqref{post_ec0bf1e2547df878846dd841b8d657b2_eq_prob_approx1} and quickly obtain a very good approximation for $P$. Consider, for instance, the eigenfunctions of the Hydrogen atom. They are specified by three quantum numbers ($n,l,m$) and always have the form: $$ \psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y^m_l(\theta,\phi) $$ where $Y^m_l(\theta,\phi)$ is the normalized spherical harmonic of degree $l$ and order $m$. The function $R_{nl}(r)$ is a real-valued function of $r$ which has the form (see [1], equations 4.73 - 4.75): $$ R_{nl}(r) = \displaystyle\frac{1}{r}\rho^{l+1} e^{-\rho} v(\rho) \label{post_ec0bf1e2547df878846dd841b8d657b2_eq_R} $$ where $\rho = r / (a_0 n)$, $v(\rho)$ is a polynomial function and $$ a_0 = \displaystyle\frac{4\pi \epsilon_0 \hbar^2}{me^2} = 0.529 \times 10^{-10}\textrm{m} $$ is the Bohr radius ($m$ is the mass of the electron). Equation \eqref{post_ec0bf1e2547df878846dd841b8d657b2_eq_R} implies $R_{nl}(r) \propto r^l$ so for every $l \gt 0$, $R_{nl}(0) = 0$. This means: $$ \psi_{nlm}(0,\theta,\phi) = \psi({\bf 0}) = 0 $$ whenever $l \neq 0$. In other words, writing $P_{nlm}$ for the value of $P$ for the eigenstate specified by the quantum numbers $(n,l,m)$, our approximation yields: $$ P_{nlm} = 0 \quad \textrm{if} \quad l \neq 0 $$ For the Hydrogen atom, the probability of finding the electron inside the nucleus can then only be nonzero if $l = 0$. Since $m = 0$ is the only allowed value of $m$ in this case ($m$ can take the values $-l,\ldots,-1,0,1,\ldots,l$), the wave functions we need to consider will have the form (see [1], table 4.2): $$ \psi_{n00}(r,\theta,\phi) = R_{n0}(r)Y^0_0(\theta,\phi) = R_{n0}(r) \displaystyle\frac{1}{\sqrt{4\pi}} \label{post_ec0bf1e2547df878846dd841b8d657b2_eq_psi_n00} $$

These are the equations for $\psi_{n00}(r,\theta,\phi)$ for $n = 1,2,3$ (obtained from [1], table 4.6, and equation \eqref{post_ec0bf1e2547df878846dd841b8d657b2_eq_psi_n00}): $$ \begin{eqnarray} \psi_{100}(r,\theta,\phi) &=& \displaystyle\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0} \nonumber\\[5pt] \psi_{200}(r,\theta,\phi) &=& \displaystyle\frac{1}{\sqrt{8\pi a_0^3}} \left(1 - \frac{1}{2}\frac{r}{a_0} \right)e^{-r/2a_0} \nonumber\\[5pt] \psi_{300}(r,\theta,\phi) &=& \displaystyle\frac{1}{\sqrt{27\pi a_0^3}} \left[1 - \frac{2}{3}\frac{r}{a_0} + \frac{2}{27}\left(\frac{r}{a_0}\right)^2\right]e^{-r/3a_0} \end{eqnarray} $$ For $r = 0$, we obtain: $$ \begin{eqnarray} \psi_{100}(0,\theta,\phi) &=& \displaystyle\frac{1}{\sqrt{\pi a_0^3}} \nonumber\\[5pt] \psi_{200}(0,\theta,\phi) &=& \displaystyle\frac{1}{\sqrt{8\pi a_0^3}} \nonumber\\[5pt] \psi_{300}(0,\theta,\phi) &=& \displaystyle\frac{1}{\sqrt{27\pi a_0^3}} \label{post_ec0bf1e2547df878846dd841b8d657b2_eq_psi_0} \end{eqnarray} $$ We can now use the values of equation \eqref{post_ec0bf1e2547df878846dd841b8d657b2_eq_psi_0} on equation \eqref{post_ec0bf1e2547df878846dd841b8d657b2_eq_prob_r0} to obtain the probability of finding the electron inside the nucleus for the states $n = 1,2,3$ with $l = m = 0$: $$ \begin{eqnarray} P_{100} &=& \displaystyle\frac{4\pi r_0^3}{3} \frac{1}{\pi a_0^3} = \displaystyle\frac{4}{3}\left(\frac{r_0}{a_0}\right)^3 \approx 9 \times 10^{-15} \nonumber\\[5pt] P_{200} &=& \displaystyle\frac{4\pi r_0^3}{3} \frac{1}{8 \pi a_0^3} = \displaystyle\frac{1}{6}\left(\frac{r_0}{a_0}\right)^3 \approx 1.1 \times 10^{-15} \nonumber\\[5pt] P_{300} &=& \displaystyle\frac{4\pi r_0^3}{3} \frac{1}{27 \pi a_0^3} = \displaystyle\frac{4}{81}\left(\frac{r_0}{a_0}\right)^3 \approx 3.3 \times 10^{-16} \end{eqnarray} $$

Therefore, for a Hydrogen atom on the ground state, the probability of finding the electron inside the nucleus is extremely small. The probability gets even smaller for higher energy states.

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