Question

The normalized ground-state wavefunction for the electron in the hydrogen atom is $$\psi(r, \theta, \phi)=\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2} e^{-r / a_{0}}$$ where $r$ is the radial coordinate of the electron and $a_{0}$ is the Bohr radius. (a) Sketch the wavefunction versus $r.$ (b) Show that the probability of finding the electron between $r$ and $r+d r$ is given by $|\psi(r)|^{2} 4 \pi r^{2} d r.$ (c) Sketch the probability versus $r$ and from your sketch find the radius at which the electron is most likely to be found. (d) Show that the wavefunction as given is normalized. (e) Find the probability of locating the electron between $r_{1}=a_{0} / 2$ and $r_{2}=3 a_{0} / 2.$ 

          The normalized ground-state wavefunction for the electron in the hydrogen atom is
$$\psi(r, \theta, \phi)=\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2} e^{-r / a_{0}}$$
where $r$ is the radial coordinate of the electron and $a_{0}$ is the Bohr radius. (a) Sketch the wavefunction versus $r.$ (b) Show that the probability of finding the electron between $r$ and $r+d r$ is given by $|\psi(r)|^{2} 4 \pi r^{2} d r.$ (c) Sketch the probability versus $r$ and from your sketch find the radius at which the electron is most likely to be found. (d) Show that the wavefunction as given is normalized. (e) Find the probability of locating the electron between $r_{1}=a_{0} / 2$ and $r_{2}=3 a_{0} / 2.$
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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The normalized ground-state wavefunction for the electron in the hydrogen atom is $$\psi(r, \theta, \phi)=\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2} e^{-r / a_{0}}$$ where $r$ is the radial coordinate of the electron and $a_{0}$ is the Bohr radius. (a) Sketch the wavefunction versus $r.$ (b) Show that the probability of finding the electron between $r$ and $r+d r$ is given by $|\psi(r)|^{2} 4 \pi r^{2} d r.$ (c) Sketch the probability versus $r$ and from your sketch find the radius at which the electron is most likely to be found. (d) Show that the wavefunction as given is normalized. (e) Find the probability of locating the electron between $r_{1}=a_{0} / 2$ and $r_{2}=3 a_{0} / 2.$
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Transcript

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00:01 So in this question we're told that the wave function for an electron in the ground state of a hydrogen atom is psi of r theta phi is 1 over root pi, 1 over a0 to the three halves, e to the minus r over a0.
00:22 So first of all, let's sketch the wave function versus r.
00:28 R, uh, psi, and it's just a negative exponential, so it just falls off.
00:35 In this kind of way and it's asymptotic with zero.
00:40 So at r equals zero, we have 1 over root pi times 1 over a nought to the three halves.
00:52 And yeah, this is what it looks like.
00:55 Part b, let's find out the probability of finding an electron.
01:00 So the probability that an electron is between r and r plus dr.
01:06 Well, let's say that the probability that the electron is in some volume v is the triple integral over that volume v with respect to the volume element of the mod square of the wave function.
01:23 Now, what's this triple integral? well, it's the integral from 0 to 2 pi, d phi, from 0 to pi d theta, and from 0 to 0 to r.
01:35 Or well, yeah, so let's say that it's from phi 1 to phi 2, from theta 1 to theta 2 d theta and from r1 to r2, dr.
01:56 And the volume element is going to be r squared sine theta, and then we've got the mod square of psi, which is 1 over pi, 1 over a0, cubed, ether minus 2r over a nought.
02:18 But you can see this only depends on r.
02:20 And that means that we can integrate out the other two coordinates.
02:25 So now let's say that v is the spherical shell between r and r plus d r.
02:38 Then phi 1 is going to be 0, phi 2 is going to be 2 pi.
02:43 Theta 1 is going to be 0, theta 2 is going to be pi.
02:47 And r1 is going to be r, r2 is going to be r plus d r.
02:54 So what we can do is we can write p rrr plus d r is equal to, so we can integrate out phi from zero to two pi.
03:06 This just gives us two pi because there's no fies in here.
03:09 So take out the pi we get a two.
03:12 So we get two over a nought cubed because i've pulled out the a nought cube as well.
03:17 And then the integral from 0 to pi, sine theta, d theta, the integral from r to r plus dr, of r squared, e to the minus 2r over a nought, dr.
03:38 This integral from 0 to pi of sine theta is just equal to 2.
03:46 So this is going to be 4 over a nought cubed.
03:51 And this integral here, because it's a, over a very short interval, it's just going to be the integrand multiplied by that interval.
03:58 So it's going to be r squared e to the minus 2r over a n a d.
04:05 But what have we got here? well this is just equal to 4 pi because we multiply this by, all we've done is multiplied the integral by 4 pi d r.
04:20 4 pi r squared d r mod psi squared.
04:23 So this is 4 pi r squared d r mod psi squared.
04:31 So that's exactly what we've got here.
04:34 So now let's sketch that probability versus r.
04:42 So prr plus dr is what we just worked out, 4r squared over a nought cubed, e to the minus 2r over a nought d r.
04:55 So let's sketch that versus r.
05:04 And it's going to start off at 0.
05:10 So the r squared is going to dominate to begin with, and then the e to the minus r is going to dominate later.
05:16 So we're going to get something like this.
05:23 And now we want to work out what this maximum point is.
05:27 So let's take a derivative.
05:31 D by dr, pr, where i'm calling, this is what i'm calling pr.
05:39 D by dr of pr is 4 over a nought cubed.
05:43 We can pull that out.
05:45 And then we can take the derivative of the r and we get 2r.
05:51 And then we can take the derivative of this e to the minus 2r over a nought.
05:55 And we get minus 2 over a nought r squared, and then we call e to the minus 2 r over a nought.
06:05 So this is 8 r, e to the minus 2 r over a nought, over a nought cubed, times r.
06:16 Oh, actually we've taken out an r, so it's going to be 1 minus r over a nought.
06:21 So we can see the derivative is zero at the beginning, like we said, because of this r, and then the derivative will be zero again when r is equal to a nought.
06:29 So that's when the derivative is zero.
06:31 And what's the probability there? well, this is going to be the probability of a 0.
06:37 So p of a 0 is 4 over a 0 e squared...
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