The hydrogen atom is composed of one proton in the nucleus...

The hydrogen atom is composed of one proton in the nucleus and one electron, which moves about the nucleus. In the quantum theory of atomic structure, it is assumed that the electron does not move in a well-defined orbit. Instead, it occupies a state known as an orbital, which may be thought of as a "cloud" of negative charge surrounding the nucleus. At the state of lowest energy, called the ground state, or Is-orbital, the shape of this cloud is assumed to be a sphere centered at the nucleus. This sphere is described in terms of the probability density function $$p(r)=\frac{4}{a_{0}^{3}} r^{2} e^{-2 r / a_{0}} \quad r \geqslant 0$$ where $a_{0}$ is the Bohr radius$\left(a_{0} \approx 5.59 \times 10^{-11} \mathrm{m}\right) .$ The integral $$P(r)=\int_{0}^{r} \frac{4}{a_{0}^{3}} s^{2} e^{-2 s / a_{0}} d s$$ gives the probability that the electron will be found within the sphere of radius $r$ meters centered at the nucleus. \begin{equation} \begin{array}{l}{\text { (a) Verify that } p(r) \text { is a probability density function. }} \\ {\text { (b) Find } \lim _{r \rightarrow \infty} p(r) . \text { For what value of } r \text { does } p(r) \text { have its }} \\ {\text { maximum value? }}\\{\text { (c) Graph the density function. }} \\ {\text { (d) Find the probability that the electron will be within the }} \\ {\text { sphere of radius } 4 a_{0} \text { centered at the nucleus. }} \\ {\text { (e) Calculate the mean distance of the electron from the }} \\ {\text { nucleus in the ground state of the hydrogen atom. }}\end{array} \end{equation}

The hydrogen atom is composed of one proton in the nucleus
and one electron, which moves about the nucleus. In the
quantum theory of atomic structure, it is assumed that the
electron does not move in a well-defined orbit. Instead, it
occupies a state known as an orbital, which may be thought
of as a "cloud" of negative charge surrounding the nucleus.
At the state of lowest energy, called the ground state, or
Is-orbital, the shape of this cloud is assumed to be a sphere
centered at the nucleus. This sphere is described in terms of
the probability density function
$$p(r)=\frac{4}{a_{0}^{3}} r^{2} e^{-2 r / a_{0}} \quad r \geqslant 0$$
where $a_{0}$ is the Bohr radius$\left(a_{0} \approx 5.59 \times 10^{-11} \mathrm{m}\right) .$ The integral
$$P(r)=\int_{0}^{r} \frac{4}{a_{0}^{3}} s^{2} e^{-2 s / a_{0}} d s$$
gives the probability that the electron will be found within the
sphere of radius $r$ meters centered at the nucleus.
\begin{equation}
\begin{array}{l}{\text { (a) Verify that } p(r) \text { is a probability density function. }} \\ {\text { (b) Find } \lim _{r \rightarrow \infty} p(r) . \text { For what value of } r \text { does } p(r) \text { have its }} \\ {\text { maximum value? }}\\{\text { (c) Graph the density function. }} \\ {\text { (d) Find the probability that the electron will be within the }} \\ {\text { sphere of radius } 4 a_{0} \text { centered at the nucleus. }} \\ {\text { (e) Calculate the mean distance of the electron from the }} \\ {\text { nucleus in the ground state of the hydrogen atom. }}\end{array}
\end{equation}

Key Concepts

Calculus Techniques - Differentiation and Limits

Techniques from calculus, such as differentiation and taking limits, play a crucial role in analyzing and understanding probability density functions. Differentiation helps in finding the maximum (mode) of a probability distribution, while limits are used to understand the behavior of the function at the boundaries of its domain. These tools are essential for both theoretical analysis and practical computations in physical sciences.

Expectation Value (Mean Distance)

The expectation value, or mean value, of a probability distribution provides the average outcome, weighted by the probability. In quantum mechanics, calculating the mean distance of an electron from the nucleus involves integrating the product of the distance and the probability density function over all space. This measure is important for describing the central tendency and spatial extent of quantum states.

Cumulative Distribution Function

The cumulative distribution function (CDF) is obtained by integrating the probability density function from a lower boundary up to a certain value. In quantum mechanics, the CDF gives the probability that the particle is found within a certain spatial region. This is useful for determining the likelihood of localization of particles within specified limits.

Probability Density Function

A probability density function (pdf) is a function that describes the relative likelihood for a random variable to occur at a given point. In quantum mechanics, the square of the wavefunction provides a pdf that gives the probability distribution of a particle's position in space. This concept is central when interpreting experimental data and predictions in statistical mechanics and quantum theory.

Normalization

Normalization is the process of scaling a function so that the total probability over its entire domain equals one, ensuring the complete probability distribution is accounted for. In the context of quantum mechanics, the wavefunction must be normalized so that the integrated probability of finding a particle anywhere in space is 100%, which is a fundamental requirement for any pdf.

Quantum Atomic Orbitals

In quantum mechanics, particles such as electrons are described by wavefunctions rather than well-defined trajectories. The concept of orbitals replaces classical orbits with probability clouds where an electron is likely to be found. This is fundamental in explaining atomic structure, chemical bonding, and the behavior of electrons in atoms.

Transcript

00:01 In this question, we are given that the hydrogen atom is composed of some other different things, and we're going to assume that it's going to be in a cloud of some sort.

00:16 But we have that the cloud will call a sphere within the nucleus, and it's described in terms of some pdf, which i wrote above.

00:27 A pdf is just a probability density function.

00:30 And we want to first show that this function, lowercase p as a function of r, is such a probability density function.

00:41 See what happens as the radius increases to infinity.

00:46 We want to find the density function, which is just the cumulative density function, and a couple of other probabilities and the mean distance.

01:02 So let's first remember, let's remember first, what a probability density function requires.

01:10 We know that a pdf, it requires that the function is positive forever, and also that the area of the curve underneath it is one.

01:29 So because of the exponential function in here, we know that the function will always be positive, in addition to the square of the radius or the distance.

01:41 So all that remains is to, all that remains is to first, integrate the function.

01:49 And we're going to be integrating this function from zero to infinity, because we're told that this probability density function has a strictly positive, has a positive radius.

02:07 So this will be the integral from zero to infinity.

02:11 We'll have four over a not cubed times r squared times e to minus 2r over a0, dr.

02:23 We can factor out the constant, of course.

02:31 And we'll have just the integral of r squared times e to the minus 2r over a knot.

02:39 So how do we integrate this per se? how will we integrate this? we're going to be using integration by parts.

02:48 So integrating by parts, we're going to have that we're going to differentiate r squared.

02:57 So this will require four entries in our column.

03:01 So the derivatives go as follows, and the integrals will go as follows.

03:10 And remember that we have to divide by the derivative, which is the same as multiplying by negative a0 over 2 times e to minus 2 r over a.

03:21 And the same thing, we'll have a not squared over 4 times e.

03:27 Minus 2r over a.

03:28 And finally, let's actually move the last row down a little bit so that we don't have things in the way.

03:36 So we'll have minus a 0 cubed over 8 times e to the minus 2r over a 0.

03:44 So multiply together each of the row, each of the diagonals, and they are part of the answer.

03:54 When we do so, we're going to have that this is equal to 4 over a0 cubed, being multiplied by negative a 0 r squared over 2 times e to minus 2r over a 0, minus 2a0 squared r over 4 times e to minus 2r over a0 and we'll have a minus 2 a 0 cubed over 8 times e to minus 2r over a 0.

04:31 And all of this will be integrated from 0 to infinity.

04:38 First off, let's do some simplifications first.

04:42 We know that this fraction here becomes a 1 -half, and we know that this fraction here becomes a 1⁄2.

04:52 One -fourth.

05:02 And so we should probably take the limit of each of these terms separately.

05:07 So let's take the limit.

05:09 So we have the limit as r goes to infinity.

05:13 We'll have just a not squared.

05:18 We'll have a not r squared over two times e to the two r over a not.

05:28 As we can see, this is an infinity over infinity, lopitol situation.

05:34 So we'll be taking the derivative.

05:37 So we have two r.

05:39 2 times r times a not over when we have the derivative here we'll have 2 times e times the something e to the something times the derivative of the something.

05:54 So we'll have 2 over a0 times e to the 2r over a not.

06:02 As we can see, when we do some simplifications, we'll have this become a 4.

06:10 That will become a 4.

06:13 And this a0 will go up to the top.

06:22 As you can see, this is still an infinity over infinity situation.

06:26 So we'll have that this is going to be equal to 2 times a not squared, divided by 4 times 2 over a0 times e to the 2r over a0.

06:42 And despite the fact that we're going to simplify this, this will actually become 0.

06:52 And i should write down the limit at each step two.

06:59 So, well, i forgot to do that here, but it's there.

07:04 So by the same logic, by the same logic for the second and the third terms, when i plug in infinity, the whole thing will become zero.

07:16 That leaves us, that leaves us with 4 over a not cubed times negative 0 minus 0 minus 0.

07:28 Subtracted from, let's plug in zero into all of the r's.

07:33 The first term and the second term will vanish, but the third term, it will not vanish, will have minus a not cubed over 4 times e to the 0, which is equal to, we'll have 4 over a not cubed times a not cubed over 4, which is just 1.

07:57 So it checks.

08:02 We verified that this is a probability density function.

08:05 Now, the second step is we want the limit as r goes to infinity of our probability density function, lowercase p of r.

08:25 This is just going to be the limit as r goes to infinity.

08:30 We'll have 4 times r cubed over a0 cubed.

08:37 I'm just rewriting this function a little bit times e to the 2r over a0...

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