Probability of finding the electron in the nucleus
Task number: 4418
An electron in the ground state of the hydrogen atom is described by the wavefunction
\[ \psi_{100}(r)=\frac{1}{\sqrt{\pi a^3}}\, \text{e}^{-\frac{r}{a}} ,\]where \( a=\frac{\hbar^2 4\pi\epsilon_0}{m_e e^2} \,\dot{=}\, 0{,}5 \,\cdot\, 10^{-11} \,\text{m}\) is the so‑called Bohr radius.
Determine the probability of finding such an electron within the region occupied by the nucleus, i.e., inside a sphere of radius \(R = 10^{-15}\, \text{m} \,.\)
Solution – beginning
The probability of finding the electron in the given region is obtained by integrating the probability density \(\vert\psi\vert^2\) over that region. For the hydrogen atom, we perform the integration in spherical coordinates.
The probability that an electron in the ground state of the hydrogen atom is located inside a sphere of radius \(R\) is
\[P=\int_0^{R} \int_0^{2\pi} \int_0^\pi |\psi_{100}(r)|^2\, r^2\, \sin\theta\, \mbox{d}\theta\, \mbox{d}\varphi\, \mbox{d}r .\]Substituting the wavefunction, we obtain
\[P= \int_0^{R} \int_0^{2\pi} \int_0^\pi \frac{1}{\pi a^3}\,\text{e}^{-\frac{2r}{a}}\, r^2\sin\theta\, \mbox{d}\theta\, \mbox{d}\varphi\, \mbox{d}r .\]Since the variables separate, this triple integral can be rewritten as a product of three simple integrals
\[P=\frac{1}{\pi a^3}\int_0^{2\pi} \mbox{d}\varphi \int_0^\pi\sin\theta\, \mbox{d}\theta \int_0^{R} r^2\,\text{e}^{-\frac{2r}{a}}\,\mbox{d}r .\]Hint – how to integrate
The calculation of the integral is described in the hint for the task Elektron mezi kulovými plochami (Czech version only).
Solution – continuation
We have found that the probability of finding the electron in the ground state of the hydrogen atom inside a sphere of radius \(R\) is given by
\[P=\frac{1}{\pi a^3}\int_0^{2\pi} \,\mbox{d}\varphi \int_0^\pi\sin\theta\,\mbox{d}\theta \int_0^{R} r^2\,\text{e}^{-\frac{2r}{a}}\,\mbox{d}r\,.\]The integrals over the angular variables are straightforward, yielding, \[P=\frac{1}{\pi a^3} \cdot 2\pi\,\cdot\,\left[-\cos \theta\right]_0^{\pi}\,\cdot\, \int_0^{R} r^2\,\text{e}^{-\frac{2r}{a}}\,\mbox{d}r\,. \] The integral over r can be evaluated by the substitution \(x=-\frac{2r}{a}\) and integration by parts (see the hint for details). After inserting the correct limits, we obtain \[P=\frac{1}{\pi a^3} \cdot 2\pi\,\cdot\,2\,\cdot\,\left(-\frac{a^3}{8}\right)\,\cdot\, \left[\left( x^2\,-\,2x\,+\,2 \right)\text{e}^x\right]_{0}^{-\frac{2R}{a}}\,,\] \[P=\frac{1}{2}\left[\left( x^2\,-\,2x\,+\,2\right)\text{e}^x\right]_{-\frac{2R}{a}}^{0}\,,\] \[P=\frac{1}{2}\left[2-\left(\left(\frac{2R}{a}\right)^2\,+\,2 \left(\frac{2R}{a}\right)\,+\,2 \right)\text{e}^{-\frac{2R}{a}}\right]\,.\]
Hint – numerical substitution
Substituting the numerical values is, in this case, beyond the numerical capabilities of a standard calculator or Excel. To obtain an approximate value of the probability, use the Taylor expansion of the exponential \(\mbox{e}^x \) and simplify the expression in the square brackets.
Solution – numerical substitution
To calculate the numerical value of the probability, we use the Taylor expansion of the exponential function around zero \(\mbox{e}^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\)
The expression \[\left( x^2\,-\,2x\,+\,2\right)\mbox{e}^x\] can then be simplified using the expansion up to the third order as \[ \begin{align*} &\left( x^2\,-\,2x\,+\,2\right) \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}\right) \approx \\ &\approx \left(x^2 + x^3 -2x -2x^2 -x^3 -2 +2x +x^2 +\frac{x^3}{3}\right) = 2 + \frac{x^3}{3} \,. \end{align*} \]
Substituting \( x=-\frac{2r}{a} \) into the probability formula, we obtain \[ \begin{align*} P \approx \frac{1}{2}\left[2-\left(2 + \frac{x^3}{3}\right)\right] = \frac{4}{3}\left(\frac{R}{a}\right)^3 \end{align*}\,. \] By inserting the numerical values \(a=5 {\cdot} 10^{-11}\) a \(R = 10^{-15}\), we obtain \[P \approx \frac{4}{3}\left(\frac{10^{-15}}{5 {\cdot} 10^{-11}}\right)^3 \approx 10^{-14} \,.\]
Answer
The probability of finding the electron inside the region occupied by the nucleus is extremely small for the ground state of the hydrogen atom, approximately \(10^{-14}\,.\)
Link
A very similar situation is also addressed in tasks Elektron mezi kulovými plochami, Elektron vně koule and Pravděpodobnost nalezení elektronu za Bohrovým poloměrem (Czech versions only).
