Answers Examination Radiation Physics - 8N120
2 November 2012 - 14.00-17.00
1. (a) The energy of photons with a wavelength of 550 nm is
Ephoton = hν =
hc
1240 eV nm
=
= 2.2545 eV .
λ
550 nm
(b) We first convert 5 Watt to electron-volt per second:
5 W = 5 J/s = 5 ∗ 6.242 · 1018 eV/s = 3.121 · 1019 eV/s
Since each photon has an energy of 2.2545 eV, there are
3.121 · 1019
= 1.384 · 1019
2.2545
photons emitted per second.
2. (a) The wavelength of the incoming photons is
λ1 =
c
hc
1240 eV nm
= 0.00124 nm .
=
=
ν1
E1
1 · 106 eV
(b) The wavelength of the scattered photons is
λ01 = λ1 +
h
h
(1 − cos(180◦ )) = λ1 + 2
= 0.00124 + 2 ∗ 0.002426 = 0.00609 nm .
me c
me c
The corresponding energy is
E10 = hν10 =
hc
1240 eV nm
= 0.204 MeV .
=
0
λ1
0.00609 nm
(c) The wavelength of the incoming photons is now
λ2 =
c
hc
1240 eV nm
=
=
= 0.000124 nm .
ν2
E2
10 · 106 eV
(d) The wavelength of the scattered photons is
λ02 = λ2 +
h
h
(1 − cos(180◦ )) = λ2 + 2
= 0.000124 + 2 ∗ 0.002426 = 0.00498 nm .
me c
me c
The corresponding energy is
E20 = hν20 =
hc
1240 eV nm
=
= 0.249 MeV .
0
λ2
0.00498 nm
(e) For incoming photons with very high energy, the wavelength λin → 0. Hence the wavelength after scattering λscat → 2 mhe c , which means that the energy of the scattered
photons becomes
Escat =
hc
hc me c
=
= 12 me c2 =
λscat
2h
1
1
2
0.511 MeV = 0.255 MeV .
3. (a) The Heisenberg uncertainty relations give that the uncertainty in the momentum is given
by
h̄
h̄
∆px =
= 9.967 · 10−25 kg m/s .
=
∆x
2 ∗ a0
The uncertainty in the velocity is then
∆vx =
∆px
= 1.094 · 106 m/s .
me
(b) The energy of the electron in the ground state of hydrogen is
E = −13.6 eV = −2.178 · 10−18 J .
Since 21 mv 2 = −E the velocity of the electron is
p
p
v = −2E/me = 2 ∗ 2.178 · 10−18 /9.109 · 10−31 = 2.187 · 106 m/s .
Note that v is much smaller than the speed of light c, hence the use of the classical
formula for the kinetic energy of the electron is justified.
4. (a) The Schrödinger equation for this potential is
−
h̄2 d2 Ψ(x) q
− Ψ(x) = EΨ(x) .
2m dx2
x
(b) The first and second derivatives of the given wave function Ψ(x) are
dΨ(x)
= Ae−ax − Aaxe−ax ,
dx
d2 Ψ(x)
= −2Aae−ax + Aa2 xe−ax .
dx2
(c) Substitution of Ψ(x) in the Schrödinger equation gives:
−
q
h̄2
(−2Aae−ax + Aa2 xe−ax ) − A x e−ax = E A x e−ax
2m
x
Rearranging terms results then in
2
Aa2 h̄2
−ax 2Aah̄
−ax
e
−
− qA + x e
− EA = 0 .
2m
2m
Since this has to hold for all positive x, we obtain the conditions
2Aah̄2
− qA
2m
−
= 0
Aa2 h̄2
− EA = 0 ,
2m
or equivalently
ah̄2
−q
m
= 0
a2 h̄2
+E = 0 .
2m
(d) The values of a and E found by solving these two equations are
a=
mq
h̄2
and E = −
a2 h̄2
mq 2
=− 2 .
2m
2h̄
(e) The value of A is found by requiring
the probability
density function Ψ2 (x) must
R ∞ that
R
∞
be normalized. This means that 0 Ψ2 (x)dx = 0 A2 x2 e−2ax dx = 1 must hold.
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(f) Using that
R∞
0
2
1
x2 e−2ax dx = (2a)
3 = 4a3 the normalization condition becomes
Z ∞
Z ∞
1
2
2
x2 e−2ax dx = A2 3 = 1 ,
Ψ (x)dx = A
4a
0
0
which implies that A2 = 4a3 , and hence
A = 2a3/2 =
2m3/2 q 3/2
.
h̄3
5. (a) The Q-value, written in terms of nuclear masses is
A−4
4
M̂
−
M̂
c2 .
Q= A
M̂
−
2
Z
Z−2
A
(b) Using that A
Z M̂ = Z M − Zme (thus omitting the relatively small electron binding energies) we obtain
A−4
A−4
4
2
A
4
Q= A
M
−
Zm
−
M
+
(Z
−
2)m
−
M
+
2m
c
=
M
−
M
−
M
c2 .
e
e
e
Z
2
Z
2
Z−2
Z−2
(c) Initially the total momentum is zero, hence after the reaction the total momentum must
also be zero. Hence the daughter A−4
Z−2 Y and the α-particle will have opposite directions
after the reaction. If the daughter has velocity w (to the left) and the α-particle has a
velocity v (to the right), momentum conservation implies that (since the weight of an
atom is approximately proportional to the mass number A):
(A − 4)v = 4w .
Squaring this yields
(A − 4)2 v 2 = 42 w2 ,
which can be rewritten in terms of the kinetic energy of the daughter TY and α-particle
Tα as:
(A − 4)TY = 4Tα .
This implies that TY =
4
A−4 Tα .
Since TY + Tα = Q, we obtain
4
Tα + Tα = Q ,
A−4
which implies
Tα =
A−4
Q
A
and
4
Q.
A
Clearly for heavy mother atoms (large A) almost all energy goes to the α-particle.
TY =
6. Consider the part of the nuclides chart in Figure 1.
(a) The Q-value of the beta minus decay
198 Au
→ 198 Hg + e− + ν is given by
198
2
Q1 = (198
79 M − 80 M )c = (197.968242 − 197.966769) ∗ 931.5 = 1.372 MeV .
The Q-value of the electron capture
198 Au
+ e− → 198 Pt + ν is given by
198
2
Q2 = (198
79 M − 78 M )c = (197.968242 − 197.967893) ∗ 931.5 = 0.325 MeV .
The Q-value of the beta plus decay
198 Au
→ 198 Pt + e+ + ν is given by
198
2
Q3 =(198
79 M − 78 M − 2me )c =
(197.968242 − 197.967893 − 2 ∗ 0.000549) ∗ 931.5 = −0.698 MeV .
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Figuur 1: part of the nuclides chart for assignment 6
(b) The beta minus decay and the electron capture have a positive Q-value, hence these
reactions are energetically possible. The beta plus decay is energetically not possible.
(c) The (right half of the) entry for 198 Au is only blue, which indicates that only beta minus
decay has been observed.
(d) The chart shows a beta minus decay of 1.0 MeV, a beta minus decay of 1.4 MeV and
a gamma decay of 412 keV. The total energy for this reaction is Q1 = 1.372 MeV. The
decay may proceed as 1) one beta minus decay of 1.4 MeV, or 2) a beta minus decay of
1.0 MeV (which leads to an excited state of 198 Hg), followed by a gamma decay of 0.412
MeV. In both cases the available energy Q1 is used.
(Since the chart gives the most frequent decays first, the beta minus decay of 1.0 MeV
occurs more often than the beta minus decay of 1.4 MeV. Thus the combination 2)
occurs the most frequently (98.99%). )
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