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Problem Set #2
1. (a) Calculate the Qvalue for K orbital-electron capture by
the nucleus,
neglecting the electron binding energy.
Ar37
18
(b) Repeat (a), including the binding energy, 3.20 keV, of the
K-shell electron in argon.
(c) What becomes of the energy released as a result of this
reaction?
2. The radioisotope224
Ra decays by emission primarily to the ground state of220
Rn
(94% probability) and to the first excited state 0.241 MeV above
the ground state (5.5%
probability). What are the energies of the two associated
particles?
3. The radionuclide41
Ar decays by emission to an excited level of
41
K that is 1.293MeV above the ground state. What is the maximum
kinetic energy of the emitted
particle?
4. The radioisotope64
Cu decays by three different mechanisms: decay (39.0%),
electron capture (EC) (43.1%) and decay (17.4 %). The Q value
for the decay is
578.7 keV. The Q value for the decay is 653.1 keV. In addition,
there is a gamma
emission (0.5% probability) at 1.345 MeV. Sketch the energy
level diagram for the
decay scheme.
+
+
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Page 2 9/28/2006
1) (# points)
a). The decay equation is:
++
CleAr37
17
0
1
37
18
Neglecting the binding energy of the electron:
QEC = parent - daughter (3.25)
= Ar Ni
= -30.951-(-31.765)= 0.814 MeV
b). Including now the electron binding energy 3.2 keV = .0032
MeV gives:
QEC = parent - daughter-EBB = 0.814-0.0032
= 0.8108 MeV
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c). The neutrino will take the entire energy, the Q energy, from
the reaction since it is practically
mass less.
2. ( # points) The decay equation for the alpha decay is:
QRnRa ++220
86
224
88
Calculating the Q value of the reaction, written with masses in
amu:
Q = Mparent Mdaughter-M (3.12)
= MRa MRn-M= 224.020202-220.011384-4.00260305
= 0.00621495 amu
= 5.789 MeV
The energy of the alpha particle is:
Mm
MQE
+
=
This is the maximum energy available for the daughter, 220Rn
going to the ground state.
Calculating the energy of the alpha particle gives:
MeVE 69.52204
)789.5(220=
+
=
The other less frequent alpha occurring 5.5% of the time leaves
the daughter nucleus in an excited
state of 0.241 MeV above the ground state. Therefore the energy
of this alpha is:
MeVE 449.5241.069.5 ==
The nuclear decay scheme that shows this is:
224Ra
5.69 MeV
5.69 MeV
(94%)
5.449 MeV
(5.5%)
220Rn
0.241 MeV
0 MeV
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3. ( # points) The decay equation is:
QKAr +++
01
41
19
41
18
Calculating the Q energy, using masses:
Q = Mparent Mdaughter= MAr MK= 40.964500-40.961825
= 0.002675 amu
= 2.492 MeV
The beta emission here leaves the daughter 41K atom in an
excited state of 1.293 MeV. Therefore
the remaining energy will go to the beta minus and the
antineutrino, hence 1.2 MeV. The
maximum energy that the emitted beta minus will have is then the
total remaining energy with the
antineutrino receiving nothing, therefore the maximum kinetic
energy is 1.2 MeV.
4. (# points) The decay scheme is:
EC
(42.6%)0 MeV
64Cu
0 MeV
(0.5%)
+
(17.4%)
EC
(0.5%)
64Cu
2mc2
64Cu
64Ni
1.345 MeV
1.6731 MeV
0.6531 MeV
-
(39%)
0.5787 MeV
64Zn
