8-11.
n2 g 2 e ! E2 / kT g 2 !(E2 ! E1 )/ kT
=
=
e
n1 g1e ! E1 / kT
g1
E2 ! E1 ) / kT
e(
T=
=
"g
g 2 n1
n #
$ = (E2 ! E1 )/ kT = ln % 2 $ 1 &
g1 n2
' g1 n2 (
E2 ! E1
10.2eV
=
= 7790 K
!5
k ln "%(g 2 / g1 )(n1 / n2 )#&
8.67 $ 10 eV / K ln 4 $ 106
(
) (
"
#
4$10!3 eV
&
!%
!
5
% 8.67$10 eV / K (300 K )&
%'
(&
8-12.
n2 g 2 !(E2 ! E1 )/ kT 3 (
=
e
= e
n1 g1
1
8-14.
cv = 3R / M
(a) Al: cv =
(b) Cu: cv =
(c) Pb: cv =
3 (1.99cal / mole K )
27.0 g / mole
)
62.5 g / mole
3 (1.99cal / mole K )
207 g / mole
= 0.155
= 0.221cal / g K
3 (1.99cal / mole K )
)
{0.215cal / g K }
= 0.0955cal / g K
{0.0920cal / g K }
= 0.0288cal / g K
{0.0305cal / g K }
The values for each element shown in brackets are taken from the
Handbook
of Chemistry and Physics and apply at 25° C.
8-17. For hydrogen: En = !
mk 2 e 4 1
13.605687
=!
eV using values of the constants
2
2
2h n
n2
accurate to six decimal places.
E1 = !13.605687eV
E2 = !3.401422eV
E2 ! E1 = 10.204265eV
E3 = !1.511743eV
E3 ! E1 = 12.093944eV
(a)
n2 g 2 !(E2 ! E1 )/ kT 8 !10.20427 / 0.02586
=
e
= e
= 4e !395 = 4 " 10!172 # 0
n1 g1
2
n3 g3 !(E3 ! E1 )/ kT 18 !12.09394 / 0.02586
= e
= e
= 9e !468 = 9 " 10!203 # 0
n1 g1
2
(b)
n2
= 0.01 = 4e !10.20427 / kT
n1
e !10.20427 / kT = 0.0025
"
!10.20427 / kT = ln (0.0025 ) = !5.99146
T=
(c)
10.20427eV
= 19, 760 K
(5.99146 ) 8.61734 " 10!5 eV K
(
)
!12.09394 / (8.61734"10!5 )(19,760 )
n3
= 9e
= 0.00742 = 0.7%
n1
B field
8-18.
m = +1
no field
!E
m=0
m = !1
E = hf
!E
Neglecting the spin, the 3p state is doubly degenerate: l = 0,1 hence, there are
two m = 0 levels equally populated.
E = hf = hc / ! = 1.8509eV
"E =
(! = 670.79nm )
ehB
= 2.315 # 10!4 eV
2me
(a) The fraction of atoms in each m-state relative to the ground state is: (Example 8-2)
n+1
= e !1.8511 / 0.02586 = e !71.58 = 10!31.09 = 8.18 " 10!32
n
n0
= 2 " e !1.8509 / 0.02586 = 2e !71.57 = 2 " 10!31.08 = 1.64 " 10!31
n
n0
= e !1.8507 / 0.02586 = e !71.56 = 10!31.08 = 8.30 " 10!32
n!1
(b) The brightest line with the B-field “on” will be the transition from the m = 0
level, the center line of the Zeeman spectrum. With that as the “standard”,
the relative intensities will be: 8.30 / 16.4 / 8.18
0.51 / 1.00 / 0.50
!
8-21. Assuming the gases are ideal gases, the pressure is given by: P =
2N E
3 V
for
classical, FD, and BE particles. PFD will be highest due to the exclusion principle, which,
in effect, limits the volume available to each particle so that each strikes the walls more
frequently than the classical particles. On the other hand, PBE will be lowest, because the
particles tend to be in the same state, which in effect, is like classical particles with a
mutual attraction, so they strike the walls less frequently.
8-23.
!=
h
=
p
h
=
2m E
h
=
2m (3kT / 2 )
h
1/ 2
(3mkT )
1/ 3
The distance between molecules in an ideal gas (V / N )
1/ 3
PV = nRT = nRT (N A N A ) = NkT ! (V / N )
1/ 3
and equating this to ! above, (kT / P )
is found from
1/ 3
= (kT / P )
h
=
1/ 2
(3mkT )
kT
h3
P
h3
5/ 2
=
and solving for T , yields: T =
P (3mkT )3 / 2
k (3mk )3 / 2
" Ph3
#
&
T =%
3/ 2
%' k (3mk ) &(
h2
TC =
2mk
8-26.
5/ 2
3
"
101kPa ) 6.63 $ 10!34 J s
(
=%
% 3 2 $ 1.67 $ 10!27 kg 1.38 $ 10!23 J / K
'
(
(
"
#
N
$
%
$& 2! (2.315 )V %'
)
)(
#
&
5/ 2
&
(
2/5
)
= 4.4 K
2/3
(Equation 8-48)
The density of liquid Ne is 1.207 g/cm3, so
(
)(
)(
)
1.207 g / cm3 6.022 ! 1023 molecules / mol 106 cm 2 / m3
N
=
= 3.601 ! 1028 / m3
V
20.18 g / mol
(6.626 # 10
"34
J s
2
)
$ 3.601 # 1028 m3 %
T=
&
'
2 20u # 1.66 # 10"27 kg / u .381 # 10"23 J / K (& 2! (2.315 ) )'
(
)(
)
2/3
= 0.895 K
20
Thus, TC at which
Ne would become a superfluid is much lower than its
freezing temperature of 24.5K.
8-28.
E =
hf
e
hf / kT
(Equation 8-60)
!1
(a) For T = 10hf / k;
(b) For T = hf / k;
hf = kT / 10 ! E =
hf = kT ! E =
(c) For T = 0.1hf / k;
hf
1 / 10
e
"1
=
kT / 10
= 0.951kT
0.1051
hf
kT
=
= 0.582kT
e " 1 1.718
1
hf = 10kT " E =
hf
10kT
=
= 4.54 # 10!4 kT
4
e ! 1 2.20 # 10
10
According to equipartition E = kT in each case.
8-29.
! hf "
CV = 3 N A k %
&
' kT (
2
e hf / kT
(e
hf / kT
2
)
)1
As T # $, hf / kT gets small and
e hf / kT ! 1 + hf / kT + L
2
! hf " (1 + hf / kT + L)
CV = 3 N A k $
# 3 N A k = 3 N A (R / N A ) = 3R
%
2
& kT '
(hf / kT )
The rule of Dulong and Petit.
8-31.
! hf "
CV = 3R #
$
% kT &
2
e hf / kT
(e
hf / kT
2
)
'1
(Equation 8-62)
At the Einstein temperature TE = hf / k,
e1
2
CV = 3R (1)
(
2
)
e1 ! 1
= 3R (0.9207 ) = 3 (8.31J / K mol )(0.9207 )
= 22.95 K / K mol = 5.48cal / K mol
8-35. Approximating the nuclear potential with an infinite square well and ignoring the
Coulomb repulsion of the protons, the energy levels for both protons and neutrons
(
are given by En = n 2 h 2
) (8mL ) and six levels will be occupied in
2
levels with 10 protons and six levels with 12 neutrons.
22
Ne , five
2
EF (protons ) =
(5) (1240MeV
2
8 (1.0078u ! 931.5MeV / u )(3.15 fm )
2
EF (neutrons ) =
2
fm )
= 516 MeV
2
(6 ) (1240MeV
fm )
2
8 (1.0087u ! 931.5MeV / u )(3.15 fm )
= 742 MeV
E (protons ) = (3 / 5 ) EF = 310 MeV
E (neutrons ) = (3 / 5 ) EF = 445MeV
As we will discover in Chapter 11, these estimates are nearly an order of
magnitude too
large. The number of particles is not a large sample.
8-36.
E1 = h 2 / 8mL2 . All 10 bosons can be in this level, so E1 (total ) = 10h 2 / 8mL2 .
8-37. (a) f FD (E ) =
=
(b) f FD (E ) =
1
e
(E ! EF / kT )
+1
1
e
(E ! EF ) / 0.1EF
+1
1
e
(E ! EF ) / 0.5 EF
+1
(Equation 8-68)
=
=
1
10(E ! EF ) / EF
e
+1
1
e
2(E ! EF ) / EF
1.00
+1
(a) T = 0.1TF
0.75
f FD
0.5
(b) T = 0.5TF
0.25
!2 EF
!1EF
0
1EF
2 EF
3EF
4 EF
8-38.
!T "
NO
# 1$ % &
N
' Tc (
3/ 2
(Equation 8-52)
!T /2"
N
(a) O # 1 $ % c &
N
' Tc (
3/ 2
!T /4"
N
(b) O # 1 $ % c &
N
' Tc (
3/ 2
!1"
= 1$ % &
'2(
3/ 2
!1"
= 1$ % &
'4(
= 0.646
3/ 2
= 0.875