PHYS2332-Modern Physics II, Winter 2015, Assignment #6
Assigned on Monday 23 March. Due on Monday 30 March.
Question 1 Dynamic Resistance of Diodes. Equation 11.9 quantifies the current as a
function of applied voltage I = I 0 ( exp ( eV / kBT ) − 1) . A) At T = 300 K, calculate
eV / kBT and exp ( eV / kBT ) for V = 1 volts and 10 volts. B) Show that at T = 300K, the
25 × 10 −3V
dynamic resistance re ≡ ( dI / dV ) ≈
, V ≡ volts , for a typical forward bias
I
voltage ( V ∼ 1 to 10V ). Hints: use the results of part A. C) For a typical applied reverse
voltage ( V < 0 ) show that re = ∞ or at least that it is a very large number. NOTE: Full
marks will be awarded only if sufficient works are shown.
A) At T = 300 K and V = 1 V,
eV / kBT = (1.6 × 10 −19 C ) (1V ) / ⎡⎣(1.381× 10 −23 J i K −1 × 300K ) ⎤⎦ = 38.6 , and
−1
exp ( eV / k BT ) = exp(38.6) = 5.9 × 1016 . At T = 300 K and V = 10 V,
eV / kBT = (1.6 × 10 −19 C ) (10V ) / ⎡⎣(1.381× 10 −23 J i K −1 × 300K ) ⎤⎦ = 386 , and
exp ( eV / k BT ) = exp(386) = a very big number. In both cases, exp ( eV / k BT ) ≫ 1 .
B) I = I 0 ( exp ( eV / kBT ) − 1) → ( dI / dV ) = ( e / kBT ) I 0 exp ( eV / kBT ) . For V ∼ 1to10V ),
exp ( eV / kBT ) ≫ 1 → I = I 0 ( exp ( eV / kBT ) − 1) ∼ I = I 0 exp ( eV / kBT ) , which gives
( dI / dV ) = ( e / kBT ) I 0 exp ( eV / kBT ) = ( e / kBT ) I .
e / kBT = (1.6 × 10 −19 C ) / ⎡⎣(1.381× 10 −23 J i K −1 × 300K ) ⎤⎦ = 38.6V −1 ∼ 40V −1 , which gives
25 × 10 −3V
I
C) For a typical reverse voltage V ∼ −1V to − 10V , exp ( eV / kBT ) ≪ 1 . From part B,
( dI / dV ) = 40V −1I → re = ( dI / dV )−1 =
( dI / dV ) = ( e / kBT ) I 0 exp ( eV / kBT ) ∼ ( 40V −1 ) I 0 exp ( eV / kBT ) , and
( 25 × 10 V )
−3
∼ ∞ , since exp ( eV / kBT ) ≪ 1 .
I 0 exp ( eV / k )
Question 2 Problem 22 chapter 11. Hint: for a large enough reverse voltage (see question
1) show that I r = −I 0 .
re = ( dI / dV ) =
−1
Start with I = I 0 ( exp ( eV / kBT ) − 1) . As shown in the previous problem (part c), for a
typical reverse voltage V ∼ −1V to − 10V , exp ( eV / kBT ) ≪ 1 , which gives
I = −I 0 = −1.05 µ A → I 0 = 1.05 × 10 −6 A .
A) Solve I = I 0 ( exp ( eV / kBT ) − 1) → V = ( kBT / e) ⎡⎣ ln ( I / I 0 ) + 1⎤⎦ , I = 140mA = 0.14A ,
V = (1.381× 10 −23 J i K −1 × 293K /1.6 × 10 −19 C ) ⎡⎣ ln ( 0.14A /1.05 × 10 −6 A ) + 1⎤⎦ = 0.32V .
This is the voltage across the diode. For the whole circuit 6V = 0.32V + IR ,
R = ( 6V − 0.32V ) / 0.14A = 40.6Ω
B) From part A) the voltage drop across the diode is 0.32V.
Question 3 Problem 17, chapter 12. NOTE: full marks will be awarded only if sufficient
work is done.
A) For a nucleus ZA X (mass M ZA X , the binding energy of the last neutron is
( )
equivalent to the energy in the nuclear reaction ZA X → A−1Z X + n ,where A−1Z X is the
daughter nucleus with one less neutron, and n is a free proton. Just like equation
12.10 the binding energy is simply the mass of the products (RHS) minus the
initial nucleus: B = ⎡⎣ M A−1Z X + mn − M ZA X ⎤⎦ c 2 . If B > 0 then the products
(
)
( )
(RHS) have higher energy than the initial isotope ZA X , and the isotope is stable
against a decay that produces A−1Z X and a free neutron.
B) Use data from appendix 8. 36 Li → 35 Li + n :
B = ⎡⎣ M
( Li ) + m
5
3
n
−M
( Li )⎤⎦ c
6
3
2
= [ 5.012540u + 1.008665u − 6.015122u ] × 931.5
16
8
O → 158 O + n :
B = ⎡⎣ M
( O) + m
15
8
n
−M
( O )⎤⎦ c
16
8
2
= [15.003065u + 1.008665u − 15.994915u ] × 931.5
207
82
Pb →
B = ⎡⎣ M
(
206
82
Pb + n :
206
82
)
Pb + mn − M
(
207
82
MeV
= 5.67MeV
u
MeV
= 15.66MeV
u
)
Pb ⎤⎦ c 2
MeV
= 6.74MeV
u
Since B > 0 in all cases, all the isotopes will not decay by neutron decay.
Question 4 Nucleation Theory of Liquid Drop.
It is known that the stable phase is the one with the lowest Gibbs free energy. For
H 2O the standard Gibb’s free energy of the liquid (water) phase is Gl = −237kJ / mol ,
and of the vapor (gas) is Gg = −228kJ / mol . The surface tension between the water and
= [ 205.974449u + 1.008665u − 206.975881u ] × 931.5
vapor phase is σ = 0.073J / m 2 , which is the unfavorable energy per unit area needed to
form an interface between the water and vapor phase. For water the molar mass is
32 × 10 −3 kg i mol −1 and the density of water is 1000kg i m −3 . A) Calculate the difference
in free energy between the water and vapor phase Δg = Gl − Gg in unit of J i m −3 . Which
phase is stable water or vapor? Explain your answer in one sentence. B) The Gibbs free
energy of formation of a spherical water droplet of radius R, is
ΔGdrop ( R ) = ( 4 / 3) π R 3Δg + 4π R 2σ . Use similar arguments as I did in class to justify the
Weizsacker model (eqn 12.20) to explain the form of ΔGdrop . C) From the form of
ΔGdrop , explain why only large liquid drops (large R) are stable. D) Calculate the critical
radius, Rc , defined as the minimum radius of a stable liquid drop. FACTS: small
raindrops in clouds are about µ m in scale, but aerosol droplets are in nm scale.
A) Δg = Gl − Gg = −9000J i mol −1 ÷ 6.023 × 10 23 particle i mol −1 = −1.5 × 10 −20 J i part −1 .
The mass of one water molecule is 18u × 1.67 × 10 −27 kg i u −1 = 3 × 10 −26 kg , which gives
Δg = −1.5 × 10 −20 J i part −1 ÷ 3 × 10 −26 kg = −5 × 10 5 J i kg −1 . Alternatively, H2O has molar
mass 18g i mol −1 → −9000J i mol −1 ÷ 18 × 10 −3 kg i mol −1 = −5 × 10 5 J i kg −1 . Water has
density ρ = 1000kg i m −3 , which gives Δg = 5 × 10 5 J i kg −1 × 1000kg i m −3 = 5 × 10 8 J i m −3
Note the severe assumption that the water droplet is pure water of ρ = 1000kg i m −3 ,
and the interface between the water and the vapor phase is infinitely thin.
B) The energy to transform H 2O from the vapor to water phase is simply Δg , which
initially is 9000 joules per mole. In part A) we convert this into unit of energy per volume
of 5 × 10 8 J i m −3 . Note that the change in Gibbs free energy Δg already includes the
energy to compress vapor to water, and that it is not necessary to take into account the
density of the vapor phase in the calculation of Δg . For a spherical drop the creation of a
drop of radius R is (4 / 3)π R 3Δg , which is negative Δg < 0 (favorable), but the creation
of the drop requires the existence of a vapor-water interface of energy 4π R 2σ , which is
positive σ > 0 (not favorable), giving ΔGdrop ( R ) = ( 4 / 3) π R 3Δg + 4π R 2σ .
C) In ΔGdrop ( R ) = ( 4 / 3) π R 3Δg + 4π R 2σ , for small drops ( R ∼ 0 ), means that R 3 ≪ R 2 ,
and the first term is negligible, and ΔGdrop ( R ∼ 0 ) ∼ 4π R 2σ > 0 , which is greater than
zero since σ > 0 . The change in the Gibb’s free energy being greater than zero means
that formation of a water drop is not stable. This is a basic law of thermodynamics.
Any such randomly formed small drops will likely decay to back to the vapor phase.
In contrast for large droplets (i.e. large R), the first term dominates R 3 ≫ R 2 ,
ΔGdrop ( large R ) ∼ ( 4 / 3) π R 3Δg < 0 , which means that large water droplets are stable.
D) At the critical radius ΔGdrop ( RC ) = 0 = ( 4 / 3) π RC3 Δg + 4π RC2σ , which gives
RC = 3σ / Δg = 3 × 0.073J i m −2 ÷ 5 × 10 8 J i m −3 = 4.4 × 10 −10 m = 4.4A , or of diameter
8.8A, which appears small, but is still greater than the 2.75 A diameter of H 2O
Question 5 Problem 26, chapter 12. NOTE: full marks will be awarded only if sufficient
work is done. You must show computational detail of at least two isotopes.
Using equation 12.20, and the data on page 446.
3 Z ( Z − 1) e2
( N − Z )2 + δ , a = 15.8MeV , a = 18.3MeV ,
B ZA X = av A − aA A 2 / 3 −
− as
v
A
5 4πε 0 r
A
as = 23.2MeV , δ = Δ (even-even), δ = 0 (even-odd, odd-even), δ = −Δ (odd-odd),
Δ = 33MeV × A −3/ 4 . Use equation 12.2, r = r0 A1/ 3 , r0 = 1.2 × 10 −15 m . Use
( )
e2
= 1.44 × 10 −9 eV im = 1.44 × 10 −15 MeV im .
4πε 0
18
6
C , A = 18, Z = 6, N = 12, even-even nuclei so Δ = 33MeV × (18 )
r = 1.2 × 10 −12 m (18 )
1/ 3
= 3.15 × 10 −15 m ,
−3/ 4
= 3.78MeV ,
B ( 186 C ) = (15.8MeV )18 − 18.3MeV (18 )
2/3
−
2
12 − 6 )
(
− 23.2MeV
+ 3.78MeV
B(
18
6
18
C = 284.4MeV − 125.7MeV − 8.3MeV − 46.4MeV + 3.78MeV = 108MeV
)
N , A = 18, Z = 7, N = 11, odd-odd nuclei so Δ = 33MeV × (18 )
18
7
r = 1.2 × 10 −12 m (18 )
1/ 3
= 3.15 × 10 −15 m ,
B ( N ) = (15.8MeV )18 − 18.3MeV (18 )
18
7
2/3
2
11− 7 )
(
− 23.2MeV
− 3.78MeV
B(
18
8
−15
3 6 ( 6 − 1) (1.44 × 10 MeV i m )
5
3.15 × 10 −15 m
18
7
−3/ 4
= −3.78MeV ,
−15
3 7 ( 7 − 1) (1.44 × 10 MeV i m )
−
5
3.15 × 10 −15 m
18
N = 284.4MeV − 125.7MeV − 11.52MeV − 20.6MeV − 3.78MeV = 123MeV
)
O , A = 18, Z = 8, N = 10, even-even nuclei so Δ = 33MeV × (18 )
B ( O ) = (15.8MeV )18 − 18.3MeV (18 )
18
8
2/3
(10 − 8 )2 + 3.78MeV
− 23.2MeV
−3/ 4
= 3.78MeV .
−15
3 8 ( 8 − 1) (1.44 × 10 MeV i m )
−
5
3.15 × 10 −15 m
18
B ( 188 O ) = 284.4MeV − 125.7MeV − 15.36MeV − 5.2MeV + 3.78MeV = 142MeV
18
10
Ne , A = 18, Z = 10, N = 8, even-even nuclei so Δ = 33MeV × (18 )
r = 1.2 × 10 −12 m (18 )
1/ 3
= 3.15 × 10 −15 m ,
B ( Ne) = (15.8MeV )18 − 18.3MeV (18 )
18
10
2/3
2
8 − 10 )
(
− 23.2MeV
+ 3.78MeV
B(
18
10
−3/ 4
= 3.78MeV ,
−15
3 10 (10 − 1) (1.44 × 10 MeV i m )
−
5
3.15 × 10 −15 m
18
Ne = 284.4MeV − 125.7MeV − 24.7MeV − 5.16MeV + 3.78MeV = 133MeV
)
Based on the B
( X ) the most stable nuclei, in order of decreasing binding energy are:
A
Z
18
O (B = 139MeV), 10
Ne (B = 133 MeV), 187 N (B = 123 MeV), 186 C (B = 108 MeV).
Looking at appendix 8 only 188 O is stable, so the result is consistent with various
hypotheses discussed in class and textbook. In particular, for light nuclei it is well known
that the most stable nuclei obey N > Z , and N ∼ Z (number of neutrons is slightly
greater than number of proton). Hence it is not surprising that the two most stable nuclei
18
are 188 O (N = 10, Z =8) and 10
Ne (N = 8, Z = 10), since for 187 N (N – Z = – 4) and 186 C (N –
Z = – 6) the difference in the number of neutrons and protons is substantial.
18
8
( )
Verification using equation 12.10 B = ⎡⎣ Nmn + ZM 1 H − M
18
6
( X )⎤⎦ c
A
Z
2
C , A = 18, Z = 6, N = 12, use data form appendix 8
MeV
= 116MeV . This is close
u
to the nuclear liquid drop model (NLDM) value of 108 MeV.
18
7 N , A = 18, Z = 7, N = 11, use data form appendix 8
B = ⎡⎣12 (1.00866u ) + 6 (1.007825 ) − 18.026760u ⎤⎦ × 931.5
B = ⎡⎣11(1.00866u ) + 7 (1.007825 ) − 18.014082u ⎤⎦ × 931.5
MeV
= 127MeV . This is close
u
to the NLDM value of 123 MeV.
18
8
O , A = 18, Z = 8, N = 10, use data form appendix 8
B = ⎡⎣10 (1.00866u ) + 8 (1.007825 ) − 17.999160u ⎤⎦ × 931.5
MeV
= 139MeV . This is quite
u
close to the NLDM value of 142 MeV.
18
10
Ne , A = 18, Z = 10, N = 8, use data form appendix 8
B = ⎡⎣ 8 (1.00866u ) + 10 (1.007825 ) − 18.005697u ⎤⎦ × 931.5
MeV
= 132MeV . This is very
u
close to the NLDM value of 133 MeV.
Although there are some differences in the calculated values, both calculations (NLDM
18
and mass comparisons) give the result that the two most stable nuclei are 188 O and 10
Ne .
In-Class data: The data I gave in class was based on edition 3 of the textbook, which is
different for which av = 14MeV , aA = 13MeV , as = 19MeV . This gives the following
values for NLDM values:
B 186 C = 252MeV − 89.3MeV − 8.3MeV − 38MeV + 3.78MeV = 120MeV
( )
B ( N ) = 252MeV − 89.3MeV − 11.52MeV − 16.9MeV − 3.78MeV = 130MeV
B ( O ) = 252MeV − 89.3MeV − 15.36MeV − 4.2MeV + 3.78MeV = 147MeV
B ( Ne) = 252MeV − 89.3MeV − 24.7MeV − 4.2MeV + 3.78MeV = 138MeV
18
7
18
8
18
10
You will not lose marks if you use this data, as long as your conclusion is logical.
Question 6 Problem 35, chapter 12. NOTE: full marks will be awarded only if sufficient
work is done.
Here it is best to follow the method of example 12.13.
52
a) for neutron decay 26
Fe → 2651Fe + n , and the disintegration energy is
Q = ⎡⎣ M
(
52
26
)
Fe − M
(
51
26
)
Fe − mn ⎤⎦ c 2
= [ 51.948117u − 50.956825u − 1.008665u ] × 931.5
MeV
= −16.2MeV
u
56
56
Since Q > 0 the energy of the 26
Fe is less than that of 2651Fe + n , and 26
Fe is stable
against neutron decay.
52
b) for proton decay 26
Fe → 2551 Mn + p , and the disintegration energy is
Q = ⎡⎣ M
(
52
26
)
Fe − M
(
51
25
)
( )
Mn − M 1 H ⎤⎦ c 2
MeV
= −7.38MeV
u
56
56
Since Q > 0 the energy of the 26
Fe is less than that of 2551 Mn + p , and 26
Fe is stable
= [ 51.948117u − 50.948216u − 1.007825u ] × 931.5
against proton decay. Finally the mass of hydrogen, M
mass of proton m p , since in decaying from
52
26 F
to
( H ) , is used instead of the
1
1
51
25 Mn ,
the number of electrons
from the isotope decreases from 26 to 25, so the extra electron in 1H partially
compensate for this.
40
For 26
Fe , there are 26 protons and 14 neutrons. Due to the large number of protons
compare to the neutrons, this nucleus would be energetically unfavorable, since the
repulsive coulomb force between the proton would exceed the attractive nuclear force
between the nucleons (neutrons and protons). In this case it may decay by proton decay.
Question 7 Problem 68, chapter 12. NOTE: full marks will be awarded only if sufficient
work is done.
A) For 24 He , the neutron number N = 2 and the proton number Z = 2 are both magic
number 2, so its abundance is not surprising. For 168 O , the neutron number N =8
and Z = 8 are both magic number 8, so its abundance is not surprising.
B) 208
82 Pb , N = 126 and Z = 82. As stated in the textbook two of the magic numbers
are 82 and 126, so its abundance is again not surprising.
40
C) Looking at appendix8, the most abundant (almost 97%) calcium is 20
Ca with N =
20 and Z = 20, where 20 is a magic number. The next two most abundant are
44
42
20 Ca (2.1%, N = 24 and Z = 20), 20 Ca (0.647%, N = 22 and Z = 20), which are
both even-even nuclei, where the number N and Z are even. The other stable
46
isotope is 20
Ca (0.004%, N = 26 and Z = 20). Noting that in figure 12.5 (page
435) light nuclei are stable only for N = Z or N slightly larger than Z, it is not
surprising that heavier calcium isotopes with N > 26 are not stable. Note that the
light (relatively speaking) even-odd isotopes (even Z and odd N), 2041Ca (N = 21
and Z = 20) and 2043Ca (N = 23 and Z = 20). These observations are in line with the
nuclear shell model that even-even isotopes are most stable, as well as the nuclear
liquid drop model that restrict stable isotopes to for N = Z or N slightly larger
than Z. In actual fact, the magic number can be derived from a harmonic oscillator
type model of the nucleus.