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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 4: Q14P (page 156)

What is the most probable value of r, in the ground state of hydrogen? (The answer is not zero!) Hint: First you must figure out the probability that the electron would be found between r and r + dr.

Short Answer

Expert verified

The most probable value of r is in the ground state of hydrogen.

Step by step solution

01

Expression of probability

The expression for the probability is given as follows,

$P = {| 蠄 |}^{2} 4 {蟺谤}^{2} dr = \frac{4}{a^{3}} e^{- 2 r / a} r^{2} dr = p (r) dr$

Here, the value of is , $\frac{4}{a^{3}} r^{2} e^{- 2 r / a}$ and the wave function in the ground state of hydrogen, $蠄 = \frac{1}{\sqrt{{蟺补}^{3}}} e^{- r / a}$ .

02

Determination of the most probable value of r

Take the derivative of $p (r) = \frac{4}{a^{3}} r^{2} e^{- 2 r / a}$ with respect to r .

role="math" localid="1657772519231" $\frac{dp (r)}{dr} = \frac{d}{dr} (\frac{4}{a^{3}} r^{2} e^{- 2 r / a}) = \frac{4}{a^{3}} [2 {re}^{- 2 r / a} + r^{2} (\frac{2}{a} e^{- 2 r / a})] = \frac{8 r}{a^{3}} e^{- 2 r / a} (1 - \frac{r}{a})$

Equate the obtained value to zero.

$\frac{dp (r)}{dr} = 0 \frac{8 r}{a^{3}} e^{- 2 r / a} (1 - \frac{r}{a}) = 0$

For the value of r equate role="math" localid="1657772026855" $(1 - \frac{r}{a})$ to zero.

$1 - \frac{r}{a} = 0 r = a$

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Most popular questions from this chapter

(a) Work out the Clebsch-Gordan coefficients for the case $s_{1} = 1 / 2, s_{2}$ =anything. Hint: You're looking for the coefficients A and Bin

$| sm 鉄� = A | \frac{1}{2} \frac{1}{2} 鉄� | s_{2} (m - \frac{1}{2}) 鉄� + B | \frac{1}{2} (- \frac{1}{2}) 鉄� | s_{2} (m + \frac{1}{2}) 鉄�$

such that $| sm 鉄�$ is an eigenstate of . Use the method of Equations 4.179 through 4.182. If you can't figure out what $S_{x}^{(2)}$ (for instance) does to $| s_{2} m_{2} 鉄�$ , refer back to Equation 4.136 and the line before Equation 4.147. Answer:

;role="math" localid="1658209512756" $A = \sqrt{\frac{s_{2} + \frac{1}{2} 卤 m}{2 s_{2} + 1}}; B = 卤 \sqrt{\frac{s_{2} + \frac{1}{2} 卤 m}{2 s_{2} + 1}}$

where, the signs are determined by $s = s_{2} 卤 1 / 2$ .

(b) Check this general result against three or four entries in Table 4.8.

In Problem4.3 you showed that $Y_{2}^{1} (胃, 蠒) = - \sqrt{15 / 8 蟺} s i n 胃 c o s 胃 e^{i 蠒}$ . Apply the raising operator to find localid="1656065252558" $Y_{2}^{2} (胃, 蠒)$ . Use Equation 4.121to get the normalization.

Use equations 4.27 4.28 and 4.32 to construct $Y_{0}^{0}, Y_{2}^{1}$ Check that they are normalized and orthogonal

The electron in a hydrogen atom occupies the combined spin and position state $R_{21} (\sqrt{1 / 3} Y_{1}^{0} 蠂_{+} + \sqrt{2 / 3} Y_{1}^{1} 蠂_{-})$

(a) If you measured the orbital angular momentum squared $(L^{2})$ , what values might you get, and what is the probability of each?

(b) Same for the component of orbital angular momentum $(L_{z})$ .

(c) Same for the spin angular momentum squared $(S^{2})$ .

(d) Same for the component of spin angular momentum $(S_{z})$ .

Let $J 鈮� L + S$ be the total angular momentum.

(e) If you measureddata-custom-editor="chemistry" $J^{2}$ , what values might you get, and what is the probability of each?

(f) Same for $J_{z}$ .

(g) If you measured the position of the particle, what is the probability density for finding it at $r$ , $胃$ , $蠒$ ?

(h) If you measured both the component of the spin and the distance from the origin (note that these are compatible observables), what is the probability density for finding the particle with spin up and at radius ?

Consider the observables $A = x^{2}$ and $B = L_{z}$ .

(a) Construct the uncertainty principle for $蟽_{A} 蟽_{B}$

(b) Evaluate $蟽_{B}$ in the hydrogen state $蠄_{n / m}$ .

(c) What can you conclude about $< xy >$ in this state?

See all solutions

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