Question

The function $$ \Psi(r)=A\left(2-\frac{Z r}{a}\right) e^{-Z r / 2 a} $$ gives the form of the quantum mechanical wavefunction representing the electron in a hydrogen-like atom of atomic number \(Z\) when the electron is in its first allowed spherically symmetric excited state. Here \(r\) is the usual spherical polar coordinate, but, because of the spherical symmetry, the coordinates \(\theta\) and \(\phi\) do not appear explicitly in \(\Psi\). Determine the value that \(A\) (assumed real) must have if the wavefunction is to be correctly normalised, i.e. the volume integral of \(|\Psi|^{2}\) over all space is equal to unity.

Short answer

A = \frac{Z^{3/2}}{2a^{3/2}}

Step-by-step solution

Understand the given equation

The quantum mechanical wavefunction is given by: \[ \Psi(r) = A\left(2-\frac{Z r}{a}\right) e^{-Z r / 2 a} \] Our goal is to normalize this wavefunction, meaning we need to find the value of A such that the integral of \(|\Psi(r)|^{2} \) over all space is equal to 1.

Set up the normalization condition

Normalization means ensuring that: \[ \int_{0}^{\infty} |\Psi(r)|^2 r^2 dr = 1 \] We include a factor of \(r^2 \) because we are integrating in spherical coordinates and must take into account the volume element, which in spherical coordinates is \( r^2 \sin \theta \ d r d \theta d \phi \).

Calculate \ \(|\Psi(r)|^2 \)

First, compute \(|\Psi(r)|^2 \): \[ |\Psi(r)|^2 = \left| A\left(2-\frac{Z r}{a}\right) e^{-Z r / 2 a} \right|^2 = A^2\left(2-\frac{Z r}{a}\right)^2 e^{-Z r / a} \]

Simplify the integral

We now substitute \(|\Psi(r)|^2 \) into the integral: \[ \int_{0}^{\infty} A^2 \left(2-\frac{Z r}{a}\right)^2 e^{-Z r / a} r^2 dr = 1 \] This integral incorporates both the normalization factor \(A^2 \) and the wavefunction terms.

Expand the integrand

Expand \(\left(2-\frac{Z r}{a}\right)^2 \) inside the integral: \[ \left(2-\frac{Z r}{a}\right)^2 = 4 - 4\frac{Z r}{a} + \left(\frac{Z r}{a}\right)^2 \] Substituting into the integral, we get: \[ \int_{0}^{\infty} A^2 \left( 4 - 4\frac{Z r}{a} + \left(\frac{Z r}{a}\right)^2\right) e^{-Z r / a} r^2 dr = 1 \]

Integrate each term separately

Split the integral into three separate integrals: \[ A^2 \left[ 4\int_{0}^{\infty} e^{-Z r / a} r^2 dr - \frac{4Z}{a}\int_{0}^{\infty} e^{-Z r / a} r^3 dr + \left(\frac{Z}{a}\right)^2 \int_{0}^{\infty} e^{-Z r / a} r^4 dr \right] = 1 \]

Evaluate the integrals using the gamma function

Recall that \(\int_{0}^{\infty} x^n e^{-\alpha x} dx = \frac{n!}{\alpha^{n+1}} \). Hence, compute each integral: \[ 1. \int_{0}^{\infty} e^{-Z r / a} r^2 dr = \frac{2!(a/Z)^{3}}{Z^3} = \frac{2a^3}{Z^3} 2. \int_{0}^{\infty} e^{-Z r / a} r^3 dr = \frac{3!(a/Z)^{4}}{Z^4} = \frac{6a^4}{Z^4} 3. \int_{0}^{\infty} e^{-Z r / a} r^4 dr = \frac{4!(a/Z)^{5}}{Z^5} = \frac{24a^5}{Z^5} \]

Substitute the evaluated integrals back

Combine the results to obtain: \[ A^2 \left[ 4\frac{2a^3}{Z^3} - 4\frac{Z}{a} \cdot \frac{6a^4}{Z^4} + \left(\frac{Z}{a}\right)^2 \cdot \frac{24a^5}{Z^5} \right] = 1 \] Simplify each term: \[ A^2 \left[ \frac{8a^3}{Z^3} - \frac{24a^3}{Z^3} + \frac{24a^3}{Z^3} \right] = 1 \Rightarrow A^2 \left[ \frac{8a^3}{Z^3} \right] = 1 \]

Solve for A

To find A, isolate it: \[ A^2 \cdot \frac{8a^3}{Z^3} = 1 \Rightarrow A^2 = \frac{Z^3}{8a^3} \Rightarrow A = \sqrt{\frac{Z^3}{8a^3}} = \frac{Z^{3/2}}{2a^{3/2}} \]

Key concepts

Quantum Mechanics

Quantum mechanics is a branch of physics that describes the behavior of particles at very small scales, such as atoms and subatomic particles. Unlike classical mechanics, which operates on Newton's laws, quantum mechanics incorporates principles like wave-particle duality and quantization.
In the context of our exercise, the wavefunction \( \Psi(r)=A\left(2-\frac{Z r}{a}\right) e^{-Z r / 2 a} \) describes the probability amplitude of finding an electron at a distance \(r\) from the nucleus in a hydrogen-like atom. The probability distribution for finding the electron is given by the square modulus of this wavefunction, \( |\Psi(r)|^{2} \).
Understanding the wavefunction's shape and ensuring it is normalized helps in predicting the behavior of electrons in atoms, making these concepts critical for quantum chemistry and atomic physics.

Hydrogen-Like Atom

A hydrogen-like atom is an atom with only one electron, similar to hydrogen. However, the nucleus can have more than one proton, making the atomic number (\(Z\)) greater than one.
Examples include singly ionized helium (He+) or doubly ionized lithium (Li2+). The Schrödinger equation for these systems yields wavefunctions that describe the allowed energy levels and spatial distribution of the electron.
In the given exercise, the wavefunction \( \Psi(r)\) describes an electron in the first allowed spherically symmetric excited state of a hydrogen-like atom. The factor \(Z\frac{r}{a}\) accounts for the potential felt by the electron due to the nuclear charge.
The normalization process involves ensuring that the total probability of finding the electron anywhere in space is one.

Spherical Symmetry

Spherical symmetry in quantum mechanics implies that a system or a function does not change when rotations are performed around a central point.
For the wavefunction \( \Psi(r)\) given in the exercise, spherical symmetry means that it depends solely on the radial distance \(r\) and not on the angular coordinates \(\theta\) and \(\phi\). This simplifies the problem significantly because we only need to integrate over the radial coordinate when normalizing the wavefunction.
Mathematically, the integral in spherical coordinates includes a volume element \(r^{2}\sin\theta \ dr \ d\theta \ d\phi\). The symmetry allows us to ignore the angular parts and focus on integrating \( \Psi(r) \) with respect to \(r\).

Normalization Condition

Normalization is a crucial concept in quantum mechanics. It ensures that the total probability of finding a particle within the entire space is always equal to one.
In our exercise, normalization requires that: \[ \int_{0}^{\infty} |\Psi(r)|^2 r^2 dr = 1 \] This integral includes the factor \(r^{2}\) from the volume element in spherical coordinates.
To normalize the wavefunction, we calculate this integral, simplify it, and solve for the normalization constant \(A\). When correctly normalized, \(A\) ensures the probability density function \( |\Psi(r)|^{2} \) integrates to unity.

Wavefunction Integral

Evaluating wavefunction integrals is essential in quantum mechanics for determining probabilities, expectation values, and normalizing wavefunctions.
In the given exercise, the integral involves the square of the wavefunction: \[ \int_{0}^{\infty} A^2 \left(2-\frac{Z r}{a}\right)^2 e^{-Z r / a} r^2 dr = 1 \] Breaking it down, we first expanded \( \left(2-\frac{Z r}{a}\right)^2 \) and split the integral into manageable parts. Each part involves a standard integral with the form \int_{0}^{\infty} r^n e^{-\alpha r} dr\, which can be solved using the Gamma function.
Combining these integrals and solving for \(A\) ensures the wavefunction is normalized, illustrating the power of integral calculus in quantum mechanics.