Question

In the ground state of a hydrogen atom, what is the probability of finding an electron anywhere in a sphere of radius (a) \(a_{0},\) or \(\left(\text { b) } 2 a_{0} ?\right.\)

Short answer

In this problem, the aim is to calculate the probability of an electron being found in a sphere of a certain radius within a hydrogen atom's ground state. This involves applying the radial distribution function and integrating it within the desired sphere to find probabilities. Detailed answer depends on the results of the integrals.

Step-by-step solution

Recalling the wave function and the Radial Distribution Function

The wave function for ground state of hydrogen atom is given by \(\psi_{100} = \frac{1}{\sqrt{\pi a_0^{3}}}e^{-r/a_0}\), and the radial distribution function is given by \(P(r) = |\psi(r)|^2 \cdot 4 \pi r^2\). This shows the distribution of electron as function of the distance to the nucleus.

Probability Calculation for a Sphere of Radius \(a_{0}\)

Now, to find probability of electron being inside a sphere of radius \(a_{0}\), the radial distribution function (RDF) must be integrated from 0 to \(a_{0}\): \(\int_0^{a_{0}} P(r)\, dr\).

Plugging RDF into the Integral to find the Probability

We substitute the expression for RDF into the integral: \(\int_0^{a_{0}} \left(4\pi r^2 \cdot \left(\frac{1}{\sqrt{\pi a_0^{3}}}e^{-r/a_0}\right)^2\right)\, dr\). Solve this integral to get the probability of finding the electron in the sphere with radius \(a_{0}\).

Repeat Steps 2 and 3 for a Sphere of Radius \(2a_{0}\)

Repeat the same process, but this time change the limit of the integral to \(2a_{0}\) instead of \(a_{0}\) to get the probability of finding the electron in the sphere with radius \(2a_{0}\)

Key concepts

Wave Function

The wave function is a fundamental concept in quantum mechanics that describes the quantum state of a particle, such as an electron in a hydrogen atom. In the ground state of a hydrogen atom, the wave function is denoted by \(\psi_{100}\). This expression characterizes the behavior and the spatial distribution of the electron around the nucleus.
For a hydrogen atom's ground state, the wave function is given by:

• \(\psi_{100} = \frac{1}{\sqrt{\pi a_0^{3}}}e^{-r/a_0}\)

The term \(a_0\) represents the Bohr radius, a fundamental physical constant approximating the average distance between the nucleus and the electron in the ground state. The exponential term \(e^{-r/a_0}\) reflects how the probability density decreases with increasing distance \(r\) from the nucleus.
Understanding the wave function is crucial because it contains all the information about the possible locations and properties of the electron in the hydrogen atom.

Radial Distribution Function

The Radial Distribution Function (RDF) provides insight into the probability of finding an electron at a particular distance from the nucleus. It effectively represents the distribution of electron density in the hydrogen atom in three-dimensional space.
The RDF is calculated from the wave function by squaring it and multiplying by the spherical volume element, resulting in:

• \(P(r) = |\psi(r)|^2 \cdot 4 \pi r^2\)

Here, \(|\psi(r)|^2\) is the probability density, and the term \(4\pi r^2\) accounts for the surface area of a sphere at radius \(r\). This function helps visualize how likely it is to find the electron at different radial distances from the nucleus.
The RDF peaks in a hydrogen atom ground state at a certain distance, indicating the radius where the electron is most probably found.

Probability Calculation

Calculating the probability of finding an electron within a certain region in space is a key application of the radial distribution function. By integrating the RDF over a specified range, we determine the likelihood of the electron being located in that area.
To find the probability that an electron is within a sphere of radius \(a_0\), we integrate the RDF from 0 to \(a_0\):

• \(\int_0^{a_{0}} P(r)\, dr\)

This involves substituting the radial distribution function into the integral, resulting in:

• \(\int_0^{a_{0}} \left(4\pi r^2 \cdot \left(\frac{1}{\sqrt{\pi a_0^{3}}}e^{-r/a_0}\right)^2\right)\, dr\)

By solving this integral, we obtain the exact probability of finding the electron within a sphere of radius \(a_0\). The process is similar for a sphere of radius \(2a_0\), by adjusting the upper limit of integration to \(2a_0\).
These calculations give us a quantitative understanding of the electron's distribution around the nucleus.

Ground State

In quantum mechanics, the ground state refers to the lowest energy state of an electron in an atom. For a hydrogen atom, the ground state is denoted as the \(n=1\) state, where \(n\) is the principal quantum number.
This state is significant because:

• It is the most stable state of the hydrogen atom.
• The electron, in this state, exhibits the lowest possible energy without external influence.

The features of the ground state include:

• A wave function \(\psi_{100}\) that defines the probability density for the electron's position.
• An electron distribution that is spherically symmetric around the nucleus.

Understanding the ground state is essential because it provides a basis for analyzing more complex states and reactions involving hydrogen and other atoms. It also plays a crucial role in spectroscopy, where transitions from the ground state to higher energy states are observed.