Question

Question: In Problem 4.43you calculated the expectation value of${\mathit{r}}^{\mathbf{s}}$in the state${\mathbf{\psi }}_{\mathbf{321}}$. Check your answer for the special cases s = 0(trivial), s = -1(Equation 6.55), s = -2(Equation 6.56), and s = -3(Equation 6.64). Comment on the case s = -7.

Short answer

for:$$\begin{gathered} s=0:1 \\ s=-1:\frac{1}{9a} \\ s=-2:\frac{1}{135a^2} \\ s=-3:\frac{1}{405a^3} \end{gathered}$$

s = -7 Or smaller that the result will be undefined due to

negative numbers! = Undefined quantity

Step-by-step solution

Given information:

Equation 6.55, 6.56, 6.64 are

Problem 4.43-(a) Construct the spatial wave function $\left(\psi \right)$for hydrogen in the state n = 3 , l=2 . m = 1 Express your answer as a function of $r,\theta ,\varphi$, and a (the Bohr radius) only - no other variables, (p,z etc.) or ,functions, ( y , v etc.), or constants, $(A,c_0$etc.), or derivatives, allowed ( $\pi$is okay, and e, and 2 , etc.).

(b) Check that this wave function is properly normalized, by carrying out the appropriate integrals over $r,\theta ,$and $\varphi$.

(c) Find the expectation value of $r^s$in this state. For what range of s (positive and negative) is the result finite?

Step 2:Check for trivial or non-trivial solution

For n = 3,l = 2,m = 1

Then:

- For s = 0:

……….[Trivial]

- For s = -1

- For s = -2:

solve further

-For s = -3

- For s = -7

$\mathrm{undentified}\equiv \infty$

Then for s = -7 or smaller that the result will be undefined due to

negative numbers! = Undefined quantity