Chapter 8
Physics 205 Solution of Home Work
Problems
8.1
Problem 8.6
Consider the original Stern-Gerlach experiment employing an atomic beam of silver, for
which the magnetic moment is due entirely to the spin of the single valence electron of the
silver atom. Assuming the magnetic field B has magnitude 0.500 T, compute the energy
difference in electron volts of the silver atoms in the two exiting beams.
Solution
The exiting beams differ in the spin orientation or the outermost atomic electron. The energy
~
difference derives from the magnetic energy of this spin in the applied field B:
~ = g −e Sz B
U = −~µs · B
2m
= gµB Bms
With g = 2 for electrons, the energy difference between the up spin (ms = 1/2) and down
spin (ms = −1/2) orientations is
∆U = gµB B = (2)(9.273 × 10−24 J/T )(0.5 T ) = 9.273 × 10−24 J = 5.80 × 10−5 eV
.
2
8.2
CHAPTER 8. PHYSICS 205 SOLUTION OF HOME WORK PROBLEMS
Problem 8.15
Spin-Orbit Energy in an Atom. Estimate the magnitude of the spin-orbit energy for an
atomic electron in the hydrogen 2p state. (Hint: From the vantage point of the moving
electron, the nucleus circles it in an orbit with radius equal to the Bohr radius for this state.
Treat the orbiting nucleus as a current in a circular wire loop and use the result from classical
electromagnetism,
2km µ
r3
for the B field at the center of loop with radius r and magnetic moment µ. Here km =
10−7 N/A2 is the magnetic constant in SI units.)
B=
Solution
The spin or the atomic electron has a magnetic energy in the field of the orbital moment
given by Equations 8.6 and 8.12 with a g-factor of 2. or
~ = 2 e Sz B = 2µB ms B
U = −~µs · B
2me
~ originates with the orbiting electron. To estimate B,
~ we adopt the
The magnetic field B
equivalent viewpoint of the atomic nucleus (proton) circling the electron, and borrow a result
~ field at the center or a circular current loop with
from classical electromagnetism for the B
radius r:
2km µ
r3
Here km is the magnetic constant and µ = iπr2 is the magnetic moment of the loop, assuming
it carries a current i. In the atomic case, we identify r with the orbit radius and the current
i with the proton charge +e divided by the orbital period T = 2πr/v. Then
B=
µ=
evr
e
=
L
2
2me
where L p
= me vr is the√orbital angular momentum of the electron. For a p electron ` = 1
and L = `(` + 1)~ = 2~, so
µ=
√
e~ √
2 = µB 2 = 1.31 × 10−23 J/T
2me
For r we take a typical atomic dimension, say 4a◦ (= 2.12 × 10−10 m) for a 2p electron, and
find
8.2. PROBLEM 8.15
3
B = 2(10−7 N/A2 )(1.31 × 10−23 J/T )/(2.12 × 10−10 m)3 = 0.276 T
Since ms is ±1/2, the magnetic energy of the electron spin in this field is
U =
=
=
=
±µB B
±(9.27 × 10−24 J/T )(0.276 T )
±2.56 × 10−24 J
±1.59 × 10−5 eV
The up spin orientation (+) has the higher energy; the predicted energy difference between
the up (+) and down (-) spin orientations is twice this figure, or about 3.18 × 10−5 eV , a
result which compares favorably with the measured value, 5 × 10−5 eV .
4
CHAPTER 8. PHYSICS 205 SOLUTION OF HOME WORK PROBLEMS
8.3
Problem 8.17
Eight identical, noninteracting particles are placed in a cubical box of sides L = 0.200 nm.
Find the lowest energy of the system (in electron volts) and list the quantum numbers of all
occupied states if
(a) the particles are electrons and
(b) the particles have the same mass as the electron but do not obey the exclusion principle.
Solution
From Equation (7.9) we have
~2 π 2 2
(nx + n2y + n2z )
2
2mL
1.054 × 10−34 )2 (π 2 )(n2x + n2y + n2z )
=
2(9.11 × 10−31)(2 × 10−10 )2
= 1.5−18 J)(n2x + n2y + n2z )
E =
= (9.4 eV )(n2x + n2y + n2z )
(a) 2 electrons per state. The lowest states have
(nx , ny , nz ) = (1, 1, 1)
⇒
E111 = (9.4 eV )(12 + 12 + 12 ) = 28.2 eV
For (nx , ny , nz ) = (1, 1, 2) or (1, 2, 1) or (2, 1, 1)
E112 =
=
=
=
Emin =
=
=
E121
E211
(9.4eV )(12 + 12 + 22 )
56.4 eV
2 × (E111 + E112 + E121 + E211 )
2(28.2 + 3 × 56.5)
394.8 eV
(b) All 8 particles go into the (nx , ny , nz ) = (1, 1, 1) state, so
Emin = 8 × E111 = 225.6 eV
8.4. PROBLEM 8.21
8.4
5
Problem 8.21
(a) Write out the electronic configuration for oxygen (Z = 8).
(b) Write out the values for the set of quantum numbers n, `, m` , and ms , for each of the
electrons in oxygen.
Solution
(a) 1s2 2s2 2p4
(b) For the two 1s electrons, n = 1, ` = 0, m` = 0, ms = ±1/2
For the two 2s electrons, n = 2, ` = 0, m` = 0, ms = ±1/2
For the four 2p electrons, n = 2, ` = 1, m` = 1, 0, −1, ms = ±1/2
6
8.5
CHAPTER 8. PHYSICS 205 SOLUTION OF HOME WORK PROBLEMS
Problem 8.23
Which electronic configuration has the lesser energy and the greater number of unpaired
spins: [Kr]4d9 5s1 or [Kr]4d10 ? identify this element and discuss Hund’s rule in this case.
(Note: The notation [Kr] represents the filled configuration for Kr.)
Solution
All spins are paired for [Kr]4d10 and two are unpaired for [Kr]4d9 5s1 . Thus Hund’s rule
would favor the latter, but for the fact that completely filled subshells are especially stable.
Thus [Kr]4d10 with its completely filled 4d subshell has the lesser energy. The element is
palladium (Pd).