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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

Construct the spin matrices $(S_{x}, S_{y} 鈥塧苍诲 S_{z})$ , for a particle of spin 1. Hint: How many eigenstates of $S_{z}$ are there? Determine the action of $S_{z}$ , $S_{+}$ , and $S_{鈭�}$ on each of these states. Follow the procedure used in the text for spin $1 / 2$ .

The (time-independent) momentum space wave function in three dimensions is defined by the natural generalization of Equation 3.54:

$桅 (p, t) = \frac{1}{\sqrt{2 蟺 h}} {鈭�}_{鈭�}^{鈭�} e^{- ipx / h} 蠄 (x, t) dx (3.54) . 蠒 (p) 鈮� \frac{1}{{(2 蟺 h)}^{3 / 2}} 鈭� e^{- i (p . r) I h} 蠄 (r) d^{3} r . (4.223) .$

(a)Find the momentum space wave function for the ground state of hydrogen (Equation 4.80). Hint: Use spherical coordinates, setting the polar axis along the direction of p. Do the 胃 integral first. Answer:

$蠄_{100} (r, 胃, 蠒) = \frac{1}{\sqrt{{蟺补}^{3}}} e^{- r / a} (4.80) . 蠒 (p) = \frac{1}{蟺} {(\frac{2 a}{h})}^{3 / 2} \frac{1}{{[1 + {(ap / h)}^{2}]}^{2 .}} (4.224) .$

(b) Check that $桅 (p)$ is normalized.

(c) Use $桅 (p)$ to calculate $< p^{2} >$ , in the ground state of hydrogen.

(d) What is the expectation value of the kinetic energy in this state? Express your answer as a multiple of $E_{1}$ , and check that it is consistent with the virial theorem (Equation 4.218).

$< T > = - E_{n}; < V > = 2 E_{n} (4.218) .$

(a) Normalize $R_{20}$ (Equation 4.82), and construct the function $蠄_{200}$ .

(b) Normalize $R_{21}$ (Equation 4.83), and construct the function.

An electron is at rest in an oscillating magnetic field

$B = B_{0} \cos (蝇迟) \hat{k}$

where $B_{0}$ and $蝇$ are constants.

(a) Construct the Hamiltonian matrix for this system.

(b) The electron starts out (at t=0 ) in the spin-up state with respect to the x-axis (that is: $蠂 (0) = 蠂_{+}^{(x)})$ . Determine $X (t)$ at any subsequent time. Beware: This is a time-dependent Hamiltonian, so you cannot get in the usual way from stationary states. Fortunately, in this case you can solve the timedependent Schr枚dinger equation (Equation 4.162) directly.

(c) Find the probability of getting $- h / 2$ , if you measure $S_{x}$ . Answer:

$\sin^{2} (\frac{{纬叠}_{0}}{2 蝇} \sin (蝇迟))$

(d) What is the minimum field $(B_{0})$ required to force a complete flip in $S_{x}$ ?

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

See all solutions

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