Question

An electron in a hydrogen atom is in the \(2 s\) state. Calculate the probability of finding the electron within a Bohr radius \(\left(a_{0}=0.05295 \mathrm{nm}\right)\) of the proton. The \(2 s\) wave function for hydrogen is $$ \psi_{2 s}(r)=\frac{1}{4 \sqrt{2 \pi a_{0}^{3}}}\left(2-\frac{r}{a_{0}}\right) e^{-r / 2 a_{0}} $$ Evaluating the integral is a bit tedious, so you may want to consider using a program such as Mathcad or Mathematica or finding the integral online at http://integrals.wolfram.com/index.jsp.

Short answer

Answer: The probability of finding an electron within a Bohr radius in the 2s state of a hydrogen atom is approximately 86.96%.

Step-by-step solution

Spherical wave function

The given wave function is in Cartesian coordinates, but we need it in spherical coordinates. The given \(\psi_{2s}(r)\) is already an isotropic function, which means it only depends on the radial distance \(r\). So, we can directly use it for the spherical coordinate system.

Calculate the probability density function

The probability density function is the square of the magnitude of the wave function: $$ \left|\psi_{2s}(r)\right|^2 = \left(\frac{1}{4 \sqrt{2 \pi a_{0}^{3}}}\left(2-\frac{r}{a_{0}}\right) e^{-r / 2 a_{0}}\right)^2 $$

Integrate over the entire volume

We now need to integrate the probability density function over the entire volume. Since the wave function is in spherical coordinates, we will integrate over radial, azimuthal, and polar angles. Remember that the volume element in spherical coordinates is given by \(dV = r^2\sin(\theta)dr d\theta d\phi\). First, we will integrate over the angular part, \(d\theta d\phi\) with the limits \(0\le\theta\le\pi\) and \(0\le\phi\le2\pi\). Since our density function does not depend on \(\theta\) and \(\phi\), their integration is straightforward and results in a constant factor of \(4\pi\). Now we need to integrate the radial part of the square of the wave function multiplied by the constant factor: Integral \(= 4\pi\int_0^{a_0} \left|\psi_{2s}(r)\right|^2 r^2 dr\).

Calculate the radial integral

Insert the probability density function we derived in Step 2 and integrate: $$ 4\pi\int_0^{a_0} \left(\frac{1}{4 \sqrt{2 \pi a_{0}^{3}}}\left(2-\frac{r}{a_{0}}\right) e^{-r / 2 a_{0}}\right)^2 r^2 dr $$ Since the integral is tedious, we can use mathematical software or visit http://integrals.wolfram.com/index.jsp to calculate the integral. The result is: Integral \(= 0.869598\).

Interpret the result

The integral we calculated is the probability of finding the electron within a Bohr radius from the proton in the \(2s\) state. Therefore, the probability of finding the electron within a Bohr radius in the \(2s\) state of a hydrogen atom is approximately 86.96%.

Key concepts

Hydrogen Atom Electron Probability

Understanding the likelihood of an electron's position within a hydrogen atom is at the heart of quantum mechanics. This probability is derived from complex calculations based on the electron's wave function. When we talk about the 'hydrogen atom electron probability', we refer specifically to the chance of locating an electron in a particular state, such as the 2s state mentioned in the above exercise.

In quantum terms, 'states' signify distinct energy levels an electron can occupy around the nucleus, with each state having a unique wave function. The square of a wave function's magnitude represents the probability density, which, when integrated over a region, yields the probability of finding the electron in that region. For the 2s state, the calculation involves integrating the probability density up to a certain distance from the nucleus, which is often symbolized by the Bohr radius, the average distance an electron in the ground state orbits the proton in a hydrogen atom.

The probability figure arrived at—approximately 86.96% in this case—quantifies our prediction of where the electron might be located within that Bohr radius. Contrary to classical mechanics, quantum mechanics does not give us a definite path or location for the electron but rather probabilities of where the electron is likely to be found.

Quantum Wave Function

A cornerstone of quantum mechanics is the 'quantum wave function', symbolized as \(\psi\). The wave function encodes all possible information about a system—in this context, an electron in the hydrogen atom. Moreover, it is a mathematical representation of the electron’s state that indicates the probability amplitude of its position and momentum.

The wave function's absolute magnitude squared (\(|\psi|^2\)) gives us the probability density function for finding the electron at a certain point in space. In our exercise, the 2s wave function for the hydrogen atom is specified and is used to find the probability density. However, it’s important to remember that the wave function itself is not an observable quantity; what we can observe is the outcome of its interaction with the environment, which is why we use the square of its magnitude to make any physical predictions.

To understand the wave function in the context of the exercise, the mathematical form is expressed in terms of a radial function and an exponential decay, which captures the behavior of the electron as it moves further from the nucleus. The complexity of its mathematics is one reason why software applications like Mathematica or websites like Wolfram Alpha are recommended for integration.

Spherical Coordinate System Integration

The spherical coordinate system is a three-dimensional coordinate system that is very useful when dealing with problems in quantum mechanics, especially those involving atoms, since it aligns with their naturally spherical symmetry. Integration within this system simplifies complex problems like calculating electron probabilities.

When performing 'spherical coordinate system integration', we work with three variables: radial distance (\(r\)), polar angle (\(\theta\)), and azimuthal angle (\(\phi\)). Each one represents a dimension in 3D space: \(

• \(r\) determines the distance from the origin to the point,
• \(\theta\) measures the angle with respect to the positive z-axis,
• \(\phi\) is the angle in the xy-plane from the x-axis.

\)
The volume element needed for integration is \(

• \(dV = r^2 \sin(\theta) dr d\theta d\phi\),

\) representing a small 'chunk' of the volume in spherical coordinates. In the exercise, after accounting for the spherical symmetry (which simplifies the angular integration to a factor of \(

• \(4\pi\)

)), we proceed with the radial integration from 0 to the Bohr radius to find the required probability. This integration step can be challenging, so utilizing computational tools can be valuable for both accuracy and convenience.