What is the probability that the 1 s electron of a hydrogen atom is found outside the Bohr radius?

The Bohr radius (a0) is a constant used in atomic physics and is the most probable distance between the proton and the electron in a hydrogen atom in its ground state (1s). It has a value of \(a_0 = 5.291772109 \times 10^{-11} \ \text{m}\).

For a hydrogen atom, the wavefunction of the 1s electron (\(\psi_{1s}\)) can be found using the Schrödinger equation. The radial part of the wavefunction (\(R_{1s}\)) has the following expression: \[R_{1s}(r) = \frac{2}{\sqrt{4\pi a_0^3}} e^{-r/2a_0}\]

To find the probability density (\(P_{1s}\)), we need to square the radial part of the wavefunction. The probability density is then given by: \[P_{1s}(r) = |R_{1s}(r)|^2 = \frac{4}{4\pi a_0^3} e^{-2r/a_0}\]

To find the probability of finding the 1s electron outside the Bohr radius, we need to integrate the probability density function over the desired region (from \(r = a_0\) to \(r = \infty\)). In spherical coordinates, the volume integral is given by: \[\text{Probability} = \int_{a_0}^{\infty} P_{1s}(r) \cdot 4\pi r^2 dr\] Now, substitute the expression for \(P_{1s}(r)\) and carry out the integration: \[\text{Probability} = \int_{a_0}^{\infty} \frac{4}{a_0^3} e^{-2r/a_0} \cdot 4\pi r^2 dr\] Using integration by substitution and integrating, we get: \[\text{Probability} = e^{-2} \approx 0.1353\] So, the probability that the 1s electron of a hydrogen atom is found outside the Bohr radius is approximately 13.53%.