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EI

# 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.

## A. we have that $p$ is a probability density funtion.B. $\lim _{r \rightarrow+\infty} p(r)=0$ and the value of $r$ at which $p$ is maximum is $r=a_{0}$C. SEE GRAPHD. $P\left(4 a_{0}\right)=1-41 e^{-8}$E. Mean distance of electron from the nucleus is $\frac{3 a_{0}}{2}$

#### Topics

Applications of Integration

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### Video Transcript

okay in this question, Russ about a probability density function P var equals for over a cubed. R squared. Eat negative two are over a For our greater than equal to zero. All right. First part asked to show that this is a P D. F. And the way you do that is by valuing the integral and seeing if it comes out to one. So let's take the integral from 0 to infinity of for over a cube S squared E. To the negative to us over a D. S. In order to do this. You're going to want to use integration by parts and I'm going to let U equal four s squared over a cubed. Do you equals eat S over a cube? Ds. D V equals E to the negative two. S over a Yes and V equals negative A over to eat a negative to us over a. So if we plug this into the formula, we'll get this equals U. V. Which is going to be negative too S squared or a squared, Eat a -2 s over a minus the integral of V. D. U. Which is going to be negative for S over a squared E. To the negative to S 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. And now this integral here is again going to be an intern creation by parts. So let's see how that will work. I'm going to have you equals for S over a squared. That makes DU equal to four over a squared D. V. Equals eat a negative two S over a D. S. So that makes V equal to negative A over to eat and negative two S over A. And for this integral by itself we're going to have UV. Which is -2 s over a eat. And they give to S over a minus the integral of VD you which is negative two over a. E. Two negative two S over A. D. S. Once again I'm gonna take this minus sign to the front. And the last integral here. This you can just evaluate with you sub that's going to come out to be plus negative E. To negative two S. Or a. Okay so now we can take all this and plug it back up here and we get that our final integral equals negative two S squared over A squared E. To negative two S over A. Then we have a minus two S over A. E. To native to S over A. And then lastly minus eat. And they give to S over a. I want to evaluate this from zero to infinity. Yeah when we plug in infinity all the exponential Xgo to zero because of the negative infinity. And the exponent. And then when we plug in zero all the S. S. Go to zero. And so the only thing that we have left is this part here Which means we're gonna have a zero he negative E. To the zeros minus one. So this does in fact equal one. And so P of our is a pdf. Okay, okay. For the next part rest for the limit as our approaches infinity of Pr and this one because P has the negative exponential blood is going to be zero. You can do this by doing low pay tolls rule or whatever method you're comfortable with. But this should make sense because we know this is a probability distribution. Eventually the probability has to go to zero. Okay. The next part says for what value of artists pr habits, maximum value. So We can solve that by saying the derivative equal to zero. So D. D. R. Of this expression for over a cubed R squared. Eat the negative to our over a. Yeah. Now we have to use the product rule. So let's do the derivative of the first. That's gonna be eight. Are a cubed needs negative to our over a plus the first time the derivative of the second, that's four over a Q R squared E two negative two are over A and using a general you get a negative two over a term there. Yeah. Now we can simplify this a little bit by factoring out a lot. We can factor out and eat and are and A to the third And an E 2 -2 are over a. What this leaves us with is the first term is just one. Second term is minus are over A. And this is all equal to zero. So this gives us our equals zero or are equals A. And it turns out that R equals A. Will give us the maximum because when you plug in R equals zero and 2 p. That just gives you zero back. So that can't be the maximum. So R equals A. Is the max here. I have graft the distribution we can see here is the maximum point. And that's going to be when R equals A. Yeah. I plotted the graph from 0-4 a. To match the next part. And so we can kind of get an idea of what percent. It's going to be in that region. Right off the bat. Yeah. Okay. Now on to party find the probability that the electron will be within the sphere of radius for a centered at the nucleus. That's gonna be the probability of for a. Which equals The integral from 0 to 4 a. Of P. S. Yes. Now, in a previous part, I found the anti derivative which is going to be the anti derivative is going to be negative too as squared over a squared E. to negative two s over a minus two S over a heat. And needed to s over a one is E. To the negative two S over A. And Instead of plugging in zero and infinity we're gonna plug into four a. So if we plug in for a here we're going to get negative. Mhm. Negative 16 over A. E. T. And the S over a. Actually that a cancer completely in the denominator. And then we have a minus eat E. 2 -2 S over a. And then we have a negative E. To negative S over a. Actually that first term Should be doubled. That's that's actually 32 here and the S. Is over. Aces can be simplified uh for a over A is just gonna be four. So all these become eat and negative eight mm. Okay. And then like before we attract a negative one So this comes one 41 E 2 -8 which is approximately 98.6%. Okay. Last part part E. It says to calculate the mean so the mean Is going to be equal to the integral from zero to infinity of S. P. M. S. Ds. So that integral, it's very similar to the integral we started off with except now we're going to have in as to the third. Just want to make sure that that is a cubed As to 3rd eat negative to S over A. Okay. And some are before we use integration by parts and when you do that you'll evaluate it out And you should get three a over two as the mean

EI

#### Topics

Applications of Integration

##### Top Calculus 2 / BC Educators ##### Heather Z.

Oregon State University   ##### Michael J.

Idaho State University

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