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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 6: Q38P (page 290)

Calculate the wavelength, in centimeters, of the photon emitted under a hyperfine transition in the ground state (n=1) of deuterium. Deuterium is "heavy" hydrogen, with an extra neutron in the nucleus; the proton and neutron bind together to form a deuteron, with spin 1 and magnetic moment
$渭_{d l} = \frac{g_{d} e}{2 m_{d}} S_{d}$
he deuteron g-factor is 1.71.

Short Answer

Expert verified

The difference in energy is $位_{d} 鈮� 92 cm$

Step by step solution

01

Definehyperfine transition.

The interaction of electrons with the magnetic moments of the nucleus splits a spectral line of an atom or molecule into two or more closely spaced components.

02

Step 2: Draw the difference in energy of deuterium.

Given

$S_{e}^{2} = \frac{1}{2} (\frac{3}{2}) {鈩�}^{2} = \frac{3}{4} {鈩�}^{2}$

渭_{d} = \frac{g_{d} e}{2 m_{d}} S_{d}, E_{h f}^{1} = \frac{渭_{0} g_{d} e^{2}}{3 蟺 m_{d} m_{e} a^{3}} <S_{d} 路 S_{e}>

Spin $S_{d}^{2} = 1 (2) {鈩�}^{2} = 2 {鈩�}^{2}$

$<S_{d} 路 S_{e}> = \{\begin{cases} \frac{1}{2} (\frac{3}{4} {鈩�}^{2} - \frac{3}{4} {鈩�}^{2} - 2 {鈩�}^{2}) = - {鈩�}^{2} \\ \frac{1}{2} (\frac{15}{4} {鈩�}^{2} - \frac{3}{4} {鈩�}^{2} - 2 {鈩�}^{2}) = \frac{1}{2} {鈩�}^{2} \end{cases}$

Difference of two state

$\frac{1}{2} {鈩�}^{2} - (- {鈩�}^{2}) = \frac{3}{2} {鈩�}^{2}$

$= \frac{g_{d} e^{2} {鈩�}^{2}}{2 蟺蔚_{0} c^{2} m_{d} m_{e} a^{3}}$

$= \frac{2 g_{d} e^{2} {鈩�}^{2}}{4 蟺蔚_{0} c^{2} m_{d} m_{e} a^{3}}$

$= \frac{2 g_{d} e^{2} {鈩�}^{4}}{c^{2} m_{d} m^{2} a^{4}}$

$= \frac{3}{2} \frac{g_{d}}{g_{p}} \frac{m_{p}}{m_{d}} 螖 E_{hydrogen}$

$位 = \frac{c}{谓} = \frac{c h}{螖 E}$

localid="1657135928063" $鈬� 位_{d} = \frac{2 g_{d}}{3 g_{p}} \frac{m_{d}}{m_{p}} 位_{h}, m_{d} = 2 m_{p}$

$鈬� 位_{d} = \frac{4}{3} (\frac{5.59}{1.71}) 路 21 cm 鈮� 92 cm$

Thus, the difference in energy is $位_{d} 鈮� 92 cm$

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Most popular questions from this chapter

Show that $P^{2}$ is Hermitian, but $P^{4}$ is not, for hydrogen states with $l = 0$ . Hint: For such states $蠄$ is independent of $胃$ and $蠒$ , so

localid="1656070791118" $p^{2} = - \frac{{鈩�}^{2}}{r^{2}} \frac{d}{d r} (r^{2} \frac{d}{d r})$

(Equation 4.13). Using integration by parts, show that

localid="1656069411605" $< f 鈭� p^{2} g > = - 鈹� 4 蟺 h^{2} (r^{2} f \frac{d g}{d r} - r^{2} g \frac{d f}{d r})_{0}^{鈭�} + < p^{2} f 鈭� g >$

Check that the boundary term vanishes for $蠄_{n 00}$ , which goes like

$蠄_{n 00} ~ \frac{1}{\sqrt{蟺 (n a)^{3 / 2}}} e x p (- r / n a)$

near the origin. Now do the same for $p^{4}$ , and show that the boundary terms do not vanish. In fact:

$< 蠄_{n 00} 鈭� p^{4} 蠄_{m 00} > = \frac{8 {鈩�}^{4}}{a^{4}} \frac{(n - m)}{(n m)^{5 / 2}} + < p^{4} 蠄_{n 00} 鈭� 蠄_{m 00} >$

Question: In the text I asserted that the first-order corrections to an n-fold degenerate energy are the eigen values of the Wmatrix, and I justified this claim as the "natural" generalization of the case n = 2.

Prove it, by reproducing the steps in Section 6.2.1, starting with

$蠄^{0} = {鈭�}_{j = 1}^{n} 伪_{j} 蠄_{j}^{0}$

(generalizing Equation 6.17), and ending by showing that the analog to Equation6.22 can be interpreted as the eigen value equation for the matrix W.

Question: In Problem 4.43you calculated the expectation value of $r^{s}$ in the state $蠄_{321}$ . 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.

When an atom is placed in a uniform external electric field ,the energy levels are shifted-a phenomenon known as the Stark effect (it is the electrical analog to the Zeeman effect). In this problem we analyse the Stark effect for the n=1 and n=2 states of hydrogen. Let the field point in the z direction, so the potential energy of the electron is

$H'_{s} = e E_{e x t} z = e E_{e x t} r c o s 胃$

Treat this as a perturbation on the Bohr Hamiltonian (Equation 6.42). (Spin is irrelevant to this problem, so ignore it, and neglect the fine structure.)

(a) Show that the ground state energy is not affected by this perturbation, in first order.

(b) The first excited state is 4-fold degenerate: $Y_{200}, Y_{211}, Y_{210}, Y_{200}, Y_{21 - 1}$ Using degenerate perturbation theory, determine the first order corrections to the energy. Into how many levels does $E_{2}$ split?

(c) What are the "good" wave functions for part (b)? Find the expectation value of the electric dipole moment $(p_{e} = - er)$ in each of these "good" states.Notice that the results are independent of the applied field-evidently hydrogen in its first excited state can carry a permanent electric dipole moment.

Consider the (eight) $n = 2$ states, $| 2 l j m_{j} 鉄�$ . Find the energy of each state, under weak-field Zeeman splitting, and construct a diagram like Figure 6.11 to show how the energies evolve as $B_{ext}$ increases. Label each line clearly, and indicate its slope.

See all solutions

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