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 $ 1s-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^3_0} r^2 e^{\frac{-2r}{a_0}} $ $ r \ge 0 $ where $ a_0 $ is the Bohr radius $ (a_0 \approx 5.59 \times 10^{-11} m) $. The integral $$ P(r) = \int_0^r \frac{4}{a^3_0} s^2 e^{\frac{-2r}{a_0}}\ ds $$ gives the probability that the electron will be found within the sphere of radius $ r $ meters centered at the nucleus. (a) Verify that $ p(r) $ is a probability density function. (b) Find $ lim_{r\to\infty} p(r) $. For what value of $ r $ does $ p(r) $ have its maximum value? (c) Graph the density function. (d) Find the probability that the electron will be within the sphere of radius $ 4_{a_0} $ centered at the nucleus. (e) Calculate the mean distance of the electron from the nucleus in the ground state of the hydrogen atom.

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 $ 1s-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^3_0} r^2 e^{\frac{-2r}{a_0}} $ $ r \ge 0 $
where $ a_0 $ is the Bohr radius $ (a_0 \approx 5.59 \times 10^{-11} m) $. The integral
$$ P(r) = \int_0^r \frac{4}{a^3_0} s^2 e^{\frac{-2r}{a_0}}\ ds $$
gives the probability that the electron will be found within the sphere of radius $ r $ meters centered at the nucleus.
(a) Verify that $ p(r) $ is a probability density function.
(b) Find $ lim_{r\to\infty} p(r) $. For what value of $ r $ does $ p(r) $ have its maximum value?
(c) Graph the density function.
(d) Find the probability that the electron will be within the sphere of radius $ 4_{a_0} $ centered at the nucleus.
(e) Calculate the mean distance of the electron from the nucleus in the ground state of the hydrogen atom.

Transcript

00:01 Okay, in this question, we're asked about a probability density function, p of r equals 4 over a cubed r squared e to negative 2r over a for r greater than equal to zero.

00:22 All right.

00:24 First part asks to show that this is a pdf, and the way you do that is by evaluating the integral and seeing a it comes out to 1.

00:34 So let's take the integral from 0 to infinity of 4 over a cubed, s squared, e to the negative, 2s over a, ds.

00:51 In order to do this, you're going to want to use integration by parts, and i'm going to let u equal 4s squared over a cubed, d u equals 8 s over a cubed d s dv equals e to the negative 2 s over a dv equals e to the negative 2 s over a d s and v equals negative a over 2 e to negative 2 s over a.

01:26 So if we plug this into the formula, we'll get this equals uv which is going to be negative 2 s squared over a squared e to the negative 2 s over a minus the integral of v du you which is going to be negative 4 s over a squared e to the negative 2s over a d s and i will take this minus sign that's in the integral and bring it out front to make this a plus.

02:11 And now this integral here is again going to be an integration by parts.

02:20 So let's see.

02:24 We're going to have u equals 4s over a squared.

02:31 That makes du equal to 4 over a squared.

02:37 Dv equals e to the negative 2 s over a d s so that makes v equal to negative a over 2 e to negative 2 s over a and for this integral by itself we're going to have uv which is negative 2 s over a a to negative 2 s over a minus the integral of v, du, which is negative 2 over a, e to negative 2s over a, ds.

03:20 Once again, i'm going to take this minus sign to the front.

03:28 And the last integral here, this you can just evaluate with u -sub.

03:36 That's going to come out to be plus negative e to negative 2s over a.

03:46 Okay.

03:51 So now we can take all this and plug it back up here and we get that our final integral equals negative 2 s squared over a squared e to negative 2 s over a then we have a minus 2 s over a e to negative 2 s over a and then lastly minus e to negative 2 s over a we want to evaluate this from zero to infinity.

04:28 When we plug it in infinity, all the exponentials go to zero, because there's a negative infinity and the exponent.

04:38 And then when we plug in zero, all the s's go to zero.

04:43 And so the only thing that we have left is this part here, which means we're going to have a zero minus negative e to the zero is minus one.

04:54 So this does, in fact, equal one.

04:57 And so p of r is a pdf.

05:04 Okay.

05:09 Okay.

05:09 For the next part, we're asked for the limit as our approach is infinity of pr.

05:21 And this one, because p has the negative exponential, the lump's going to be zero.

05:32 You can do this about doing lopi toll's rule or whatever method you're comfortable with.

05:38 But this should make sense because we know this is a probability distribution.

05:43 Eventually the probability has to go to zero.

05:46 Okay.

05:48 The next part says, for what value of r does pr have its maximum value? so we can solve that by saying the derivative equal to zero.

05:59 So ddr of this expression for, over a cubed, r squared, e to the negative to r over a.

06:14 Now we have to use the product rule.

06:17 So let's do the derivative of the first.

06:20 That's going to be 8r a cubed, each negative to r over a, plus the first times the derivative of the second...

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