Question

For hydrogen atoms, the wave function for the state \(n=3\) \(\ell=0, m_{\ell}=0\) is $$\psi_{300}=\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta}$$ where \(\sigma=r / a_{0}\) and \(a_{0}\) is the Bohr radius \(\left(5.29 \times 10^{-11} \mathrm{m}\right) .\) Calculate the position of the nodes for this wave function.

Short answer

The nodes for the wave function \(\psi_{300}\) are the positions in terms of r where the probability density of finding an electron is zero. To find these positions, we set the wave function equal to zero and solve for r. We find the node positions to be: \[r_1 = \left(\frac{18 + \sqrt{216}}{4}(5.29 \times 10^{-11}\mathrm{m})\right)\] \[r_2 = \left(\frac{18 - \sqrt{216}}{4}(5.29 \times 10^{-11}\mathrm{m})\right)\]

Step-by-step solution

Identify the wave function for the given quantum numbers

We are given the wave function for the hydrogen atom in the state with quantum numbers n=3, l=0, and m_l=0: \[\psi_{300}=\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta}\] where \(\sigma = r/a_0\) and \(a_0\) is the Bohr radius (\(5.29 \times 10^{-11}\) m).

Set the wave function equal to zero

To find the nodes, we want to find the values of r for which the wave function is equal to zero: \[\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta} = 0\]

Identify the factors affecting the result

The exponential term \(e^{-\sigma \beta}\) is always positive and can never be equal to zero. The same goes for the constants in the expression. Therefore, the key term to focus on is the polynomial one: \[27-18 \sigma+2 \sigma^{2} = 0\] Replace \(\sigma\) with \(r/a_{0}\) to find the nodes in terms of r: \[27-18\left(\frac{r}{a_{0}}\right) + 2 \left(\frac{r}{a_{0}}\right)^2 = 0\]

Solve the quadratic equation for r

Solve the quadratic equation to find the roots (i.e., node positions) in terms of r: \[2\left(\frac{r}{a_{0}}\right)^2 - 18\left(\frac{r}{a_{0}}\right) + 27 = 0\] Let \(x = \frac{r}{a_{0}}\), so the equation becomes: \[2x^2 - 18x + 27 = 0\] Use the quadratic formula to solve for x: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where a = 2, b = -18, and c = 27. \[x = \frac{18 \pm \sqrt{(-18)^2 - 4(2)(27)}}{2(2)}\] Simplify and solve for x: \[x = \frac{18 \pm \sqrt{216}}{4}\] This gives us two values for \(x\): \[x_1 = \frac{18 + \sqrt{216}}{4}\] \[x_2 = \frac{18 - \sqrt{216}}{4}\]

Convert the x values back to r-values

Recall that \(x = \frac{r}{a_{0}}\). Multiply both sides by \(a_{0}\) to obtain the node positions, \(r_1\) and \(r_2\): \[r_1 = x_1a_0 = \left(\frac{18 + \sqrt{216}}{4}\right)(5.29 \times 10^{-11}\mathrm{m})\] \[r_2 = x_2a_0 = \left(\frac{18 - \sqrt{216}}{4}\right)(5.29 \times 10^{-11}\mathrm{m})\] Now you have found the position of the nodes for this wave function.

Key concepts

Quantum Numbers

Quantum numbers are fundamental to understanding the behavior of electrons in atoms. They describe the state of an electron in a hydrogen atom and are derived from the solutions to Schrödinger's equation. In our current exercise, the hydrogen atom is described by specific quantum numbers:

• Principal Quantum Number (n): This number is n=3 in our problem. It represents the energy level of the electron and how far the electron is from the nucleus.
• Azimuthal Quantum Number (ℓ): Our exercise has ℓ=0. This number represents the shape of the electron's orbital. In this case, it's a spherical s-orbital.
• Magnetic Quantum Number (m_ℓ): This is m_ℓ=0 in our exercise and it dictates the orientation of the orbital in space. Since it's 0, the orbital orientation is straightforward, being spherically symmetric.

These numbers help characterize the electron’s wave function and thereby define its probable location around the nucleus.

Bohr Radius

The Bohr radius, denoted as \(a_0\), represents the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state. It is an important constant in quantum mechanics, with approximately the value \(5.29 \times 10^{-11}\) meters. In the context of our problem, \(a_0\) appears as a scaling factor that converts dimensionless quantities into physically meaningful distances. The variable \(\sigma = r / a_0\) used in the wave function equation allows us to work with non-dimensional terms, which simplifies the mathematical operations.The Bohr radius is foundational in defining atomic units, a convenient system of natural units used in atomic physics. Understanding the Bohr radius helps students grasp the scales at which quantum effects start to dominate.

Wave Function Nodes

Wave function nodes are points where the probability density of finding an electron is zero. In the wave function we are examining, nodes appear wherever the radial part equals zero. Nodes can be found by setting the polynomial part of the wave function to zero, as shown in:\[27 - 18 \sigma + 2 \sigma^2 = 0\]In this problem, solving this equation gives us the positions of the nodes using the variable \(\sigma\). We need to remember to solve for \(r\) by substituting \(\sigma = r/a_0\) once the nodes are found in terms of \(\sigma\).Nodes are a key concept because they affect the shape and energy of atomic orbitals. The number of nodes increases with energy levels, providing insights into how probability clouds change with different quantum states.

Quadratic Equation in Quantum Mechanics

In this exercise, solving for the positions of nodes involves unraveling a quadratic equation. This is a common scenario in quantum mechanics wherein equations of the form \(ax^2 + bx + c = 0\) frequently arise. For our wave function, rearranging terms yields:\[2\left(\frac{r}{a_{0}}\right)^2 - 18\left(\frac{r}{a_{0}}\right) + 27 = 0\]We substitute and solve for \(x = \frac{r}{a_0}\) using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]With appropriate substitutions, this resolves into solutions that describe the node positions in terms of \(r\). This process exemplifies how polynomial equations help us deduce crucial information about atomic structure and electron positioning.Solving these equations illustrates a recurring strategy in quantum theory, showing how algebra integrates with physics to reveal the fundamental behavior of matter at microscopic scales.