PS1, WI24, CHEM 126B 16. The hydrogen atom 2s orbital has...

PS1, WI24, CHEM 126B 16. The hydrogen atom 2s orbital has the following normalized wavefunction: $\psi = \left(\frac{1}{32\pi a_0^3}\right)^{1/2} \left(2 - \frac{r}{a_0}\right)e^{-r/2a_0}$ a) Calculate the probability of finding an electron in the 2s orbital in a volume of 1.5 pm³ (i) centered on the nucleus, (ii) at the Bohr radius, and (iii) at twice the Bohr radius. Hint: Assume the volume 1.5 pm³ is so small such that the wavefunction is constant in this region.

Transcript

00:01 In this question, we are given the normalized ground state wave function for the electron in the hydrogen atom.

00:06 Okay, and then in part a, we want to sketch the wave function versus r.

00:11 So i use excel.

00:15 The sketch is over here.

00:21 Okay.

00:24 Alright, so then we can move on to part b.

00:31 Okay, in part b, we want to show that the probability of finding the electron between r and r plus dr is given by given by s .r square times 4 pi r squared at the r.

00:46 Okay, so one thing we need to note is that the the brown states way function is spirically symmetric.

01:09 So, so, so sigh r, what square is also so it's very clearly symmetric.

01:28 So when we calculate the probability, when we calculate the probability, is of finding the electron between r and r plus the r.

02:00 So this is going to be equal to the probability density, the probability density, the probability density.

02:08 Okay okay which is the side r square okay mod side r square and then you multiply by the volume of the shell of the spherical shell okay so um so it's going to be uh side r square and then the volume of the spherical shell is 4 pi r square the r so shown okay right because of the because the probability density is spherically symmetric so the the value is the same at the distance are in the spherical shell so you can just multiply by the volume and to find a probability okay, okay.

03:33 Okay, then in part c, we need to sketch the probability versus r...

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