Answer
The probability of finding the electron:
$P\left( {\rho \ge {a_0}} \right) \simeq 0.677$
Work Step by Step
We have the wave function for the $1s$ state of an electron in the hydrogen atom:
${\Psi _{1s}}\left( \rho \right) = \frac{1}{{\sqrt {\pi {a_0}^3} }}{{\rm{e}}^{ - \rho /{a_0}}}$,
where ${a_0}$ is the Bohr radius.
The probability density function for the electron in spherical coordinates is given by
$p\left( \rho \right) = |{\Psi _{1s}}\left( \rho \right){|^2} = \frac{1}{{\pi {a_0}^3}}{{\rm{e}}^{ - 2\rho /{a_0}}}$
The region where the distance is greater than the Bohr radius is described in spherical coordinates by
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|{a_0} \le \rho < \infty ,0 \le \phi \le \pi ,0 \le \theta \le 2\pi } \right\}$
We evaluate the probability of finding the electron at a distance greater than the Bohr radius in spherical coordinates:
$P\left( {\rho \ge {a_0}} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} p\left( \rho \right){\rm{d}}V$
$ = \frac{1}{{\pi {a_0}^3}}\mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \mathop \smallint \limits_{\rho = {a_0}}^R {{\rm{e}}^{ - 2\rho /{a_0}}}{\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{1}{{\pi {a_0}^3}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\phi } \right)\left( {\mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{\rho = {a_0}}^R {{\rm{e}}^{ - 2\rho /{a_0}}}{\rho ^2}{\rm{d}}\rho } \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = 2\pi $ and $\mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\phi = - \left( {\cos \phi |_0^\pi } \right) = 2$, we get
$P\left( {\rho \ge {a_0}} \right) = \frac{4}{{{a_0}^3}}\left( {\mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{\rho = {a_0}}^R {{\rm{e}}^{ - 2\rho /{a_0}}}{\rho ^2}{\rm{d}}\rho } \right)$
Write $u = - 2\rho /{a_0}$. So, $\rho = - \frac{1}{2}u{a_0}$ and $d\rho = - \frac{{{a_0}}}{2}du$.
We get ${\rho ^2}{\rm{d}}\rho = - \frac{1}{8}{a_0}^3{u^2}{\rm{d}}u$. Thus,
$P\left( {\rho \ge {a_0}} \right) = - \frac{1}{2}\left( {\mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{u = - 2}^{ - 2R/{a_0}} {u^2}{{\rm{e}}^u}{\rm{d}}u} \right)$
Using the reduction formula (Exercise 60 of Section 8.1):
$\mathop \smallint \limits_{}^{} {x^n}{{\rm{e}}^x}{\rm{d}}x = {x^n}{{\rm{e}}^x} - n\mathop \smallint \limits_{}^{} {x^{n - 1}}{{\rm{e}}^x}{\rm{d}}x$
we get
$P\left( {\rho \ge {a_0}} \right) = - \frac{1}{2}\left( {\mathop {\lim }\limits_{R \to \infty } \left( {\left( {{u^2}{{\rm{e}}^u}|_{ - 2}^{ - 2R/{a_0}}} \right) - 2\mathop \smallint \limits_{u = - 2}^{ - 2R/{a_0}} u{{\rm{e}}^u}{\rm{d}}u} \right)} \right)$
$ = - \frac{1}{2}\left( {\mathop {\lim }\limits_{R \to \infty } \left( {4\frac{{{R^2}}}{{{a_0}^2}}{{\rm{e}}^{ - 2R/{a_0}}} - 4{{\rm{e}}^{ - 2}} - 2\mathop \smallint \limits_{u = - 2}^{ - 2R/{a_0}} u{{\rm{e}}^u}{\rm{d}}u} \right)} \right)$
Once again using the reduction formula, we get
$ = - \frac{1}{2}\left( {\mathop {\lim }\limits_{R \to \infty } \left( {4\frac{{{R^2}}}{{{a_0}^2}}{{\rm{e}}^{ - 2R/{a_0}}} - 4{{\rm{e}}^{ - 2}} - 2\left( {u{{\rm{e}}^u}|_{ - 2}^{ - 2R/{a_0}} - \mathop \smallint \limits_{u = - 2}^{ - 2R/{a_0}} {{\rm{e}}^u}{\rm{d}}u} \right)} \right)} \right)$
$ = - \frac{1}{2}(\mathop {\lim }\limits_{R \to \infty } \left( {4\frac{{{R^2}}}{{{a_0}^2}}{{\rm{e}}^{ - 2R/{a_0}}} - 4{{\rm{e}}^{ - 2}} - 2\left( { - 2\frac{R}{{{a_0}}}{{\rm{e}}^{ - 2R/{a_0}}} + 2{{\rm{e}}^{ - 2}} - {{\rm{e}}^u}|_{ - 2}^{ - 2R/{a_0}}} \right)} \right)$
$ = - \frac{1}{2}\left( {\mathop {\lim }\limits_{R \to \infty } \left( {4\frac{{{R^2}}}{{{a_0}^2}}{{\rm{e}}^{ - 2R/{a_0}}} - 4{{\rm{e}}^{ - 2}} + 4\frac{R}{{{a_0}}}{{\rm{e}}^{ - 2R/{a_0}}} - 4{{\rm{e}}^{ - 2}} + 2{{\rm{e}}^{ - 2R/{a_0}}} - 2{{\rm{e}}^{ - 2}}} \right)} \right)$
$ = - \frac{1}{2}\left( {\mathop {\lim }\limits_{R \to \infty } \left( {4\frac{{{R^2}}}{{{a_0}^2}}{{\rm{e}}^{ - 2R/{a_0}}} + 4\frac{R}{{{a_0}}}{{\rm{e}}^{ - 2R/{a_0}}} + 2{{\rm{e}}^{ - 2R/{a_0}}} - 10{{\rm{e}}^{ - 2}}} \right)} \right)$
$ = - \frac{1}{2}\left( {\mathop {\lim }\limits_{R \to \infty } \left( {4\frac{{{R^2}}}{{{a_0}^2{{\rm{e}}^{2R/{a_0}}}}} + 4\frac{R}{{{a_0}{{\rm{e}}^{2R/{a_0}}}}} + \frac{2}{{{{\rm{e}}^{2R/{a_0}}}}} - 10{{\rm{e}}^{ - 2}}} \right)} \right)$
Since the exponentials in the denominators grow faster than those terms in the numerators, we have
$\mathop {\lim }\limits_{R \to \infty } \left( {4\frac{{{R^2}}}{{{a_0}^2{{\rm{e}}^{2R/{a_0}}}}} + 4\frac{R}{{{a_0}{{\rm{e}}^{2R/{a_0}}}}} + \frac{2}{{{{\rm{e}}^{2R/{a_0}}}}}} \right) = 0$
So, $P\left( {\rho \ge {a_0}} \right) = 5{{\rm{e}}^{ - 2}}$
Hence, $P\left( {\rho \ge {a_0}} \right) = \frac{5}{{{e^2}}} \simeq 0.677$.
