Question

The radial wave function for hydrogen in the \(1 s\) state is given by \(R_{1 s}=A_{1} e^{-r / a_{0}},\) where the normalization constant, \(A_{1},\) was found in Example 38.2 a) Calculate the probability density at \(r=a_{0} / 2\). b) The 1 s wave function has a maximum at \(r=0\) but the 1 s radial probability density peaks at \(r=a_{0}\). Explain this difference.

Short answer

Answer: The probability density at \(r = a_{0} / 2\) for the \(1s\) state of hydrogen is \(P(a_{0}/2) = \frac{\pi a_{0}^2}{e^1}\). The wave function has a maximum at \(r=0\) due to the highest value of the radial wave function, while the radial probability density has a maximum at \(r=a_{0}\), as it depends on the product of the surface area of a sphere and the probability amplitude, which is greatest at that point.

Step-by-step solution

Calculate the probability amplitude with the given function

To find the probability amplitude, we first need to find the square of the radial wave function. The probability amplitude is given by \(|R_{1s}(r)|^2 = (A_{1} e^{-r / a_{0}})(A_{1} e^{-r / a_{0}}) = A_{1}^2 e^{-2r / a_{0}}\).

Calculate the probability density at r = a_{0}/2

The probability density is given by multiplying the probability amplitude by the volume element (\(P(r)dr = |R(r)|^2\cdot 4\pi r^2 dr\)). Therefore, \(P(r) = |R_{1s}(r)|^2\cdot 4\pi r^2\). At \(r = a_{0}/2\), we have \(P(a_{0}/2) = A_{1}^2 e^{-2a_{0}/2a_{0}}\cdot 4\pi (a_{0}/2)^2 = A_{1}^2 e^{-1}\cdot 4\pi \frac{a_{0}^2}{4}\). By simplifying, we get \(P(a_{0}/2) = A_{1}^2 e^{-1} \pi a_{0}^2\). According to Example 38.2, \(A_1 = \sqrt{\frac{1}{\pi a_0^3}}\). Thus, the probability density at \(r=a_{0}/2\) is \(P(a_{0}/2) = \frac{\pi a_{0}^2}{e^1}.\)

Explain the difference between the maximums

The wave function has a maximum at \(r=0\) because the value of the radial wave function is at its highest when \(r=0\). On the other hand, the radial probability density depends not only on the radial wave function but also on the volume element, which is determined by the surface area of a sphere (\(4\pi r^2\)). Since we have a factor \(r^2\) in the volume element, this causes the radial probability density to be zero at \(r=0\). The radial probability density peaks at \(r=a_{0}\) because the product of the surface area of a sphere and the probability amplitude is greatest at that point.