Problem 19 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Show that the mean value of \(r\) for an electron in the ground state of hydrogen is \(\bar{r}=\frac{3}{2} r_{0},\) by calculating $$ \overline{\boldsymbol{r}}=\int_{\text {all space }} \boldsymbol{r}\left|\psi_{100}\right|^{2} d \boldsymbol{V}=\int_{0}^{\infty} \boldsymbol{r}\left|\psi_{100}\right|^{2} 4 \boldsymbol{\pi} r^{2} d r $$

Step-by-Step Solution

Answer

The mean value \(\bar{r}\) is \(\frac{3}{2} a_0\).

Step 1: Recall the Ground State Wave Function

The ground-state wave function for an electron in hydrogen (quantum numbers \(n=1\), \(l=0\), \(m=0\)) is given by \( \psi_{100}(r,\theta,\phi) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} \), where \(a_0\) is the Bohr radius. This wave function is spherically symmetric and only depends on the radial coordinate \(r\).

Step 2: Set Up the Integral for Mean Position

The mean value of \(r\) is determined by the integral \(\bar{r} = \int_{0}^{\infty} r |\psi_{100}(r)|^2 4\pi r^2 \ dr\). Substitute \(|\psi_{100}(r)|^2 = \left(\frac{1}{\pi a_0^3} e^{-2r/a_0}\right)^2\) into this expression.

Step 3: Simplify the Expression Under the Integral

Substitute \(|\psi_{100}(r)|^2 = \left(\frac{1}{\pi a_0^3} e^{-r/a_0}\right)^2\) into the integral. This gives us: \[ \bar{r} = \int_{0}^{\infty} r \left(\frac{1}{\pi a_0^3} e^{-2r/a_0}\right) 4\pi r^2 \ dr = \frac{4}{a_0^3} \int_{0}^{\infty} r^3 e^{-2r/a_0} \ dr \]

Step 4: Evaluate the Integral

This is a standard integral of the form \(\int_{0}^{\infty} x^n e^{-ax} \ dx = \frac{n!}{a^{n+1}}\) with \(n=3\) and \(a=2/a_0\). Plugging in the values, we get: \[ \int_{0}^{\infty} r^3 e^{-2r/a_0} \ dr = \frac{3!}{(2/a_0)^4} = \frac{6}{16/a_0^4} = \frac{3}{8} a_0^4 \]

Step 5: Compute the Mean Value

Substitute the evaluated integral back into the expression for \(\bar{r}\) to find:\[ \bar{r} = \frac{4}{a_0^3} \cdot \frac{3a_0^4}{8} = \frac{4 \cdot 3a_0}{8} = \frac{3}{2} a_0 \] This shows that the mean value of \(r\) for the electron in the ground state of hydrogen is \(\bar{r} = \frac{3}{2} a_0\).

Step 6: Conclusion

Thus, we have proven that the mean radial distance for an electron in the ground state of hydrogen is \(\frac{3}{2} a_0\).

Key Concepts

Hydrogen Atom

A hydrogen atom is the simplest atom and consists of just one proton and one electron. In quantum mechanics, it serves as an excellent model system for studying atomic structure because it has only one electron. This makes its mathematical treatment more straightforward compared to multi-electron atoms.
The electron in a hydrogen atom moves in a probabilistic manner around the nucleus, described by a wave function. Hydrogen is also the element where quantum mechanics and classical physics were first reconciled effectively, particularly with Bohr's model. This model allows us to understand energy levels and atomic spectra for hydrogen, paving the way for modern quantum mechanics.

Wave Function

In quantum mechanics, the wave function is a crucial concept that describes the quantum state of a particle. For a hydrogen atom, this wave function provides the probability of finding an electron in various locations around the nucleus. It is a mathematical expression with three quantum numbers: principal ( egin{math} egin{align*} n eq 0 eq 0 eq 0 eq 0 n eq 0 eq 0 n eq 0 n eq 0 n eq 0 n eq 0 n eq 0 n eq 0 nnn eq 0 eq 0 eq 0 nnnn eq 0 * 0 eq 0 eq 0 nnn n nnnnn n eq 0 eq 0 nn n nnnnn* 0 eq 0 nnnnn Probability: 0 eq 0 Wave functions are keys to probability; they describe where electrons could be. nnnnnn n nnnnn* 0 n n

Bohr Radius

The Bohr radius, denoted as \(a_0\), is a fundamental physical constant in the Bohr model of the atom. It represents the most probable distance of the electron from the nucleus in a hydrogen atom when it is in its ground state. The Bohr radius is significant because it sets the scale of atomic dimensions and is approximately equal to 0.529 angstroms (~0.529 .
Using the Bohr radius, one can express other quantum characteristics, such as the energy levels and radii of orbits for electrons. In quantum mechanics, the Bohr radius is crucial because it provides a natural unit of length, allowing the atom's scale to be understood in a simple and uniform manner.

Ground State

The ground state of an atom refers to the lowest energy state that an atom can have. For a hydrogen atom, this means that its single electron is in the lowest allowed energy level, with principal quantum number \(n = 1\). This state is the most stable configuration, as the atom has the least amount of energy.
When an atom is in its ground state, the probability distribution of the electron is given by the ground state wave function. This function is spherically symmetric, reflecting that the electron is most likely to be found near the Bohr radius. The ground state is important because it is the reference state when evaluating energy differences during transitions to excited states, forming the basis for phenomena like spectral lines.