Question

The radial wave function for hydrogen in the \(1 s\) state is given by \(R_{1 s}=A_{1} e^{-r / a_{0}}\) a) Calculate the normalization constant \(A_{1}\). b) Calculate the probability density at \(r=a_{0} / 2\). c) The \(1 s\) wave function has a maximum at \(r=0\) but the \(1 s\) radial density peaks at \(r=a_{0} .\) Explain this difference.

Short answer

#Answer#: The difference between the maximum of the 1s wave function and the peak of the 1s radial density lies in accounting for the volume element in the spherical coordinate system. The 1s wave function has a maximum at r=0, whereas the radial density peak occurs at r=a_0 due to the inclusion of the volume element (\(4 \pi r^2 dr\)), which is zero at r=0.

Step-by-step solution

Finding the normalization condition for the radial wave function

First, we need to find the normalization condition for the radial wave function \(R_{1s}\). In general, the normalization condition \(N(r)\) for a radial wave function is given by: \(N(r) = \int_{0}^{\infty}|R_{1s}(r)|^{2} 4 \pi r^{2} dr = 1 \)

Integrating the normalization condition

Now, we need to integrate the normalization condition, N(r), to find the constant A_1. Plugging in the given radial wave function function \(R_{1s}=A_{1} e^{-r / a_{0}}\), we get: \(\int_{0}^{\infty}|A_{1} e^{-r / a_{0}}|^{2} 4 \pi r^{2} dr = 1\) Now, do the integral with respect to r: \(\int_{0}^{\infty}(A_{1})^{2} e^{-2r / a_{0}} 4 \pi r^{2} dr = 1\)

Solving the integral and calculating A_1

Solving the integral as shown in step 2 and then finding A_1, we get: \((A_{1})^{2} \cdot \int_{0}^{\infty} e^{-2r / a_{0}} 4 \pi r^{2} dr = 1\) Solving this integral gives: \((A_{1})^{2} \cdot \frac{a_{0}^{3} \pi}{4} = 1\) Now, we can find the normalization constant \(A_{1}\): \(A_{1} = \sqrt{\frac{4}{a_{0}^{3}\pi}}\)

Calculating the probability density at r=a_0 / 2

Next, let's find the probability density at r=a_0 / 2. The probability density is given by the squared value of the radial wave function as follows: \(P(r) = |R_{1s}(r)|^{2}\) Plug the found normalization constant A_1, \(R_{1s}= \sqrt{\frac{4}{a_{0}^{3}\pi}} e^{-r / a_{0}} \), and r = a_0/2: \(P \left(\frac{a_0}{2}\right) = \left|\sqrt{\frac{4}{a_{0}^{3}\pi}} e^{-\frac{1}{2}}\right|^{2}\) \(P \left(\frac{a_0}{2}\right) = \frac{4}{a_{0}^{3}\pi} e^{-1}\)

Explaining the difference between the 1s wave function maximum and the radial density peak

The 1s wave function has a maximum at \(r=0\) because the exponential term in the radial wave function \(R_{1s}=A_{1} e^{-r / a_{0}}\) has a maximum at r=0. However, the radial density is given by multiplying the wave function by r^2, which is zero at r=0. Therefore, the peak in the radial density occurs at a different point, which is \(r=a_0\), where the product of r^2 and the wave function is maximum. In simpler terms, the 1s wave function maximum occurs where the wave function is maximum, but the radial density takes into account the volume element in the spherical coordinate system (\(4 \pi r^2 dr\)).

Key concepts

Quantum Mechanics and the Hydrogen Atom

Quantum mechanics provides a fundamental theoretical framework for understanding the behavior of particles at atomic and subatomic levels. One of its central features is the description of electrons in an atom using wave functions. For the hydrogen atom, these functions describe the probability amplitude of finding the electron at a certain position.

The hydrogen atom is one of the simplest and most studied systems in quantum mechanics. It consists of a single proton and an electron, which allows for relatively straightforward solutions to the Schrödinger equation - the equation that governs the behavior of quantum systems. The radial wave function for hydrogen, denoted by R_{nl}(r), where n is the principal quantum number and l is the angular momentum quantum number, describes the radial part of an electron's wave function.

In the case of the 1s state, which refers to the lowest energy state (ground state) of hydrogen with zero angular momentum, the wave function is spherically symmetric, meaning it only depends on the distance r from the nucleus. It's important to note that the radial wave function must be normalized, ensuring the total probability of finding the electron anywhere in space is one.

Probability Density in Quantum Systems

The probability density is a measure that represents how likely it is to find a particle, such as an electron, at a particular location in a quantum system. Mathematically, it is the square of the modulus of the wave function (|ψ(r)|^2), which for the radial part becomes |R_{nl}(r)|^2.

In hydrogen's ground state, or 1s state, the radial wave function tells us how the electron's presence probability density changes with the distance r from the nucleus. The probability density is highest at the nucleus (r=0) and decreases exponentially with distance. However, when considering the radial probability density, which combines the probability density with the spherical volume element 4πr^2, the peak is found not at the nucleus but at a certain radius r=a_0 where a_0 is the Bohr radius. This accounts for the fact that an electron's probability to be located is distributed throughout a volume, and not solely at a point.

Wave Function Normalization

Normalization is a critical concept in quantum mechanics that ensures the wave function describes a particle with certainty to be found somewhere in space. To normalize a wave function, the integral of the probability density over all space must equal one. For the spherical symmetry in the hydrogen atom's 1s state, the integral to compute normalization involves the radial probability density and the volume element in spherical coordinates.

The normalization constant, A_1, for the hydrogen atom's radial wave function ensures the total integrated probability density equals one. The process involves setting up an integral of the square of the radial wave function times the spherical volume element and solving for A_1. This ensures that, when we calculate specific probabilities, such as finding an electron within a certain radius, our values will be accurate representations of the electron's behavior in the 1s state of a hydrogen atom.

Illustrating Normalization with the Hydrogen Atom

To normalize the hydrogen atom's 1s radial wave function, one must solve an integral that takes into account the exponential decay of the probability amplitude with distance from the nucleus, alongside the increase in volume element as r grows. The result gives us a concrete value for A_1, which is then used to calculate other properties, such as the probability density at specific radii.