What is the probability that an electron in the ground state of hydrogen will be found inside the nucleus?

1. First calculate the exact answer, assuming the wave function is correct all the way down to$\mathbf{r}\mathbf{=}\mathbf{0}$. Let b be the radius of the nucleus.
2. Expand your result as a power series in the small number$\mathit{a}\mathbf{=}\mathbf{2}\mathit{b}\mathbf{}\mathit{l}\mathbf{}\mathit{a}$, and show that the lowest-order term is the cubic:$\mathbf{P}\mathbf{\approx }\left(4l3\right){\left(bla\right)}^{\mathbf{3}}$. This should be a suitable approximation, provided that$\mathbf{b}\mathbf{\ll }\mathbf{a}$(which it is).
3. Alternatively, we might assume that$\mathbf{\psi }\left(r\right)$is essentially constant over the (tiny) volume of the nucleus, so that$\mathit{P}\mathbf{\approx }\left(4l3\right)\mathit{\pi }{\mathit{b}}^{\mathbf{3}}\mathbf{}\mathit{l}\mathit{\psi }\left(0\right)\mathbf{}{\mathit{l}}^{\mathbf{2}}$.Check that you get the same answer this way.
4. Use$\mathbf{b}\mathbf{\approx }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{}\mathbf{m}$and${\mathbf{a}}^{\mathbf{\approx }\mathbf{0}}\mathbf{\approx }\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{m}$to get a numerical estimate for$\mathbf{P}$. Roughly speaking, this represents the fraction of its time that the electron spends inside the nucleus:"

Hydrogen atoms ground state is described by${\mathbf{\psi }}_{\mathbf{100}}\left(r,\theta ,\varphi \right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{\pi a}}^{\mathbf{3}}}}{\mathbf{e}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{a}}$

(a)

The probability of finding the electron inside a nucleus with radius is:

$$\begin{gathered} P=\int {\overline{)\psi }}^2{d}^{3}r \\ =\frac{1}{\pi a^3}{\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^b{e}^{-2rla{r}^2}\sin \left(\theta \right)drd\theta d\varphi \\ =\frac{4\pi }{\pi a^3}{\int }_0^b{e}^{-2rla{r}^2}dr \\ ={\overline{)\frac{4}{a^3}\left[-\frac{a}{2}{r}^2{e}^{-2rla}+\frac{a^3}{4}{e}^{-2rla}\left(-\frac{2r}{a}-1\right)\right]}}_0^b \\ =-{\overline{)\left(1+\frac{2r}{a}+\frac{2r^2}{a^2}\right)e^{-2rla}}}_0^b \\ =1-\left(1+\frac{2b}{a}+2\frac{2b^2}{a^2}\right)e^{-2rla} \\ P=1-\left(1+\frac{2b}{a}+2\frac{2b^2}{a^2}\right)e^{-2rla} \end{gathered}$$

Thus,

the probability is$1-\left(1+\frac{2b}{a}+2\frac{2b^2}{a^2}\right)e^{-2rla}$.

(b)

Assume $\epsilon =2b/a$in the result of part (a), and get,

$$\begin{gathered} P=1-\left(1+\in +\frac{1}{2}{\in }^2\right)e^{-\in } \\ P=1-\left(1+\in +\frac{1}{2}{\in }^2\right)\left(1-\in \frac{{\in }^2}{2}-\frac{{\in }^3}{3!}\right) \\ P=1-\left(1+\in +\frac{1}{2}{\in }^2\right)\left(1-\in \frac{{\in }^2}{2}-\frac{{\in }^3}{3!}\right) \\ =1-1+\in -\frac{\in 2}{2}+\frac{{\in }^7}{6}-\in +{\in }^2-\frac{\in 2}{2}-\frac{\in 2}{2}+\frac{\in 2}{2} \\ ={\in }^3\left(\frac{1}{6}-\frac{1}{2}+\frac{1}{1}\right) \\ P=\frac{4}{3}{\left(\frac{b}{a}\right)}^3 \end{gathered}$$

Thus, the lowest order term is cubic.

(c)

Alternatively, let the wave function is constant across the nucleus' volume.

$$\begin{gathered} a\approx 0.5\times {10}^{-10}m \\ =\frac{4}{3}\frac{\pi b^3}{\pi a^3} \\ =\frac{4}{3}{\left(\frac{b}{a}\right)}^3 \\ P=\frac{4}{3}{\left(\frac{b}{a}\right)}^{\mathbf{3}} \end{gathered}$$

Therefore, the result matches with the result of part b.

(d)

Calculate the probability by using$b\approx {10}^{-15}$and$a\approx 0.5\times {10}^{-10}$

$$\begin{gathered} P=\frac{4}{3}{\left(\frac{{10}^{-15}}{0.5\times {10}^{-10}}\right)}^3 \\ =\frac{4}{3}{\left(2\times {10}^{-5}\right)}^3 \\ =\frac{4}{3}.8\times {10}^{-15} \\ =\frac{32}{3}\times {10}^{-15} \\ P1.07\times {10}^{-14} \end{gathered}$$

Hence, the estimate for P is$1.07\times {10}^{-14}$.