The normalized radial wave function for the 2 p state of the...

The normalized radial wave function for the $2 p$ state of the hydrogen atom is $R_{2 p}=\left(1 / \sqrt{24 a^{5}}\right) r e^{-r / 2 a}$. After we average over the angular variables, the radial probability function becomes $P(r) d r=\left(R_{2 p}\right)^{2} r^{2} d r .$ At what value of $r$ is $P(r)$ for the $2 p$ state a maximum? Compare your results to the radius of the $n=2$ state in the Bohr model.

Key Concepts

Quantum Numbers and Orbital Structure

The 2p state of the hydrogen atom is characterized by a principal quantum number n = 2 and an orbital angular momentum quantum number l = 1. These quantum numbers uniquely determine the electron's energy and angular distribution. The quantum mechanical treatment of these states reveals that the 2p orbital has a radial wave function that exhibits a specific node structure and a peak in its radial probability distribution, reflecting the underlying principles of quantum mechanics and orbital angular momentum.

Maximization of the Radial Probability Distribution

Finding the maximum of the radial probability distribution involves taking the derivative of P(r) with respect to r and setting it equal to zero. For the 2p state, this process identifies the r value at which the function r^4 e^(-r/a) (arising from the squared radial wave function multiplied by r^2) is maximized. Through this derivation, one finds that the maximum occurs at r = 4a, where a is the Bohr radius. This peak indicates the most probable distance from the nucleus at which the electron in the 2p orbital is likely to be found.

Radial Probability Distribution

In quantum mechanics, the radial probability distribution function is defined as P(r) = [R(r)]^2 r^2, where R(r) is the radial part of the wave function. This function gives the probability density for finding the electron at a distance r from the nucleus after averaging over all angular variables. It combines the behavior of the wave function with the geometry of three-dimensional space, emphasizing how the volume element (r^2 dr) affects the probability interpretation.

Bohr Model of the Hydrogen Atom

The Bohr model provides an early quantum description of the hydrogen atom, where electrons orbit the nucleus in circular paths with quantized radii. For the n = 2 energy level, the model predicts an electron orbit radius of 4a, which corresponds exactly to the radius where the maximum probability is found in the quantum mechanical description of the 2p state. This agreement highlights a key point of correspondence between the Bohr model and the more complete quantum mechanical model in terms of characteristic distance scales for the electron in a hydrogen atom.

Transcript

00:01 To find maximum or minimum values of a function, you take the derivative of that function with respect to a variable and set it equal to zero.

00:07 So our function, the probability function, p, i replaced our function r with the value of r squared, since it's r squared times little r squared.

00:20 So i replaced that with our value of big r, the hydrogen atom 2p state wave function, and then simplified it.

00:25 So this is r to the 4th, e to the minus r over a, divided by 24 times a to the 5th.

00:30 So we take the derivative of that with respect to r and then set it equal to 0.

00:34 And then we can solve for r.

00:36 So we have dpdr.

00:41 So using the chain rule, this is equal to 4 times r to the third, or excuse me, the product rule, e to the minus r over a, divided by 24a to the 5th.

01:06 Then we have minus.

01:08 We have r to the fourth, and we're going to bring down the derivative of negative r over a, which is 1 over a.

01:15 So this is minus r to the fourth over a times e to the minus r over a divided by 24 times a to the fifth.

01:36 And again, you have maximum and minimums when this derivative is equal to zero.

01:43 So we can multiply both sides by 24a to the fifth, right? and zero times anything is still zero, so the 24a to the fifth cancels.

01:53 And then we can divide both sides by e to the minus r over a because that's common in both...

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