Question

An electron in a hydrogen atom is in the ground state (1s). Calculate the probability of finding the electron within a Bohr radius \(\left(a_{0}=0.05295 \mathrm{nm}\right)\) of the proton. The ground state wave function for hydrogen is: \(\psi_{1 s}(r)=A_{1 s} e^{-r / a_{0}}=e^{-r / a_{0}} / \sqrt{\pi a_{0}^{3}}\).

Short answer

Answer: The probability of finding the electron within a Bohr radius of the proton in the ground state of a hydrogen atom is approximately 86.5%.

Step-by-step solution

Understand the wave function and probability density function

The wave function, \(\psi_{1s}(r)\), gives us information about the electron in the ground state of the hydrogen atom. The probability density function is given by the square of the wave function, i.e., \(|\psi_{1s}(r)|^2\). In this case, it is \(|A_{1s} e^{-r/a_0}|^2 = |e^{-r/a_0}/\sqrt{\pi a_0^3}|^2\).

Set up the integral for the probability

In order to find the probability of finding the electron within a Bohr radius of the proton, we need to integrate the probability density function over the volume within the Bohr radius. Since we are dealing with a spherical volume, we should use spherical coordinates to set up the integral. The integral will be set up as: \(P(r \leq a_0) = \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{a_0} |\psi_{1s}(r)|^2 r^2 \sin{\theta} dr d\theta d\phi\)

Calculate the integral

Now we can plug in the wave function and evaluate the integral: \(P(r \leq a_0) = \int_{0}^{2\pi}d\phi \int_{0}^{\pi}\sin{\theta} d\theta \int_{0}^{a_0} \left(\frac{e^{-r/a_0}}{\sqrt{\pi a_0^3}}\right)^2 r^2 dr\) First, let's solve the integration in \(\theta\) and \(\phi\): \(\int_{0}^{2\pi}d\phi = 2\pi\), \(\int_{0}^{\pi}\sin{\theta} d\theta = [-\cos{\theta}]_{0}^{\pi} = 2\). Then, we find the integral in r: \(\int_{0}^{a_0} \left(\frac{e^{-r/a_0}}{\sqrt{\pi a_0^3}}\right)^2 r^2 dr = \frac{1}{\pi a_0^3}\int_{0}^{a_0} e^{-2r/a_0} r^2 dr\). Use integration by parts with \(u=r^2\) and \(dv=e^{-2r/a_0}dr\) as: \(du = 2r dr\), \(v = -\frac{a_0}{2}e^{-2r/a_0}\). We get: \(\frac{1}{\pi a_0^3}\left[-\frac{a_0}{2}r^2 e^{-2r/a_0}|_{0}^{a_0} - \frac{a_0}{2}\int_{0}^{a_0}2r e^{-2r/a_0} dr\right] = \frac{1}{\pi a_0^3}\left[-\frac{a_0}{2} a_0^2 e^{-2} + a_0^2\left(1-e^{-2}\right)\right]\) Now we find the total probability by multiplying the results of the three integrals: \(P(r \leq a_0) = (2\pi)(2)\left(\frac{1}{\pi a_0^3}\left[-\frac{a_0}{2} a_0^2 e^{-2} + a_0^2\left(1-e^{-2}\right)\right]\right) = 4\left(1-e^{-2}\right) \approx 0.865\) So, there is an approximately 86.5% chance of finding the electron within a Bohr radius of the proton.

Key concepts

Ground State Wave Function

The ground state wave function in quantum mechanics represents the electron's behavior when it occupies the lowest energy level. For the hydrogen atom, this is described by the equation \(\psi_{1s}(r) = \frac{e^{-r/a_0}}{\sqrt{\pi a_{0}^{3}}}\), where \(a_0\) is the Bohr radius, and \(r\) is the distance from the proton. The wave function is crucial because it provides a way to calculate the probability of an electron's position in space.

Bohr Radius

The Bohr radius (\(a_0\)) is approximately 0.0529 nm and is a fundamental physical constant that describes the size of the orbit in which the electron resides in the hydrogen atom. In terms of quantum mechanics, it acts as a scale factor in the wave function and represents the most likely distance between the electron and the nucleus in the ground state.

Probability Density Function

The probability density function is derived from the square of the wave function and it describes how dense the probability of finding an electron is at various points around the nucleus. Mathematically, for the ground state wave function of hydrogen, it is \(\psi_{1s}(r)^2\). By integrating this function over a specific volume, we determine the likelihood of locating the electron within that region.

To find this probability within one Bohr radius, as seen in the exercise, you would set up an integral over the space of interest, which in this case is a sphere with radius \(a_0\).

Quantum Mechanics

Quantum mechanics forms the theoretical basis that explains the nature and behavior of matter and energy on the atomic and subatomic level. Unlike classical physics, where objects have definite positions and velocities, quantum mechanics adopts the concept of probabilities and wave functions to predict where particles like electrons could be found.

This branch of physics introduces a radically different framework to comprehend the interactions of particles and is essential for calculating the behavior of electrons within atoms, such as those in the hydrogen atom's ground state.

Spherical Coordinates

Spherical coordinates are a system for expressing the position of points in three-dimensional space using three numbers: the radial distance, the polar angle, and the azimuthal angle. These coordinates are particularly useful for problems involving symmetry around a point, such as the electronic probability density of an atom.

When computing the probability of an electron's location around the atom, we use spherical coordinates to integrate the probability density function over a sphere's volume, as is demonstrated in the step-by-step solution to finding the likelihood of the electron being within one Bohr radius.