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SACE Stage 2 Physics
Radioactivity 1 Solutions
1.
(i)
For Z> 82 nuclides have too many protons for stability, as a consequence
they decay by emitting particles. This reduces the number of protons by 2.
At the same time it also reduces the number of neutrons. (the consequence
of the protons undergoing positron decay to neutrons
( 1p1  on1 + 1o + 00  ) would be to shift the nuclide well away from the
stability line as a result of a growing number of neutrons in the nucleus.
This is not energetically possible.
(ii)
Consequence of proton
decay would be to shift the
nuclide to the left and away
from line of stability
N
N=Z
Stability line
Z
As it is the  decay moves the nuclide down parallel to the Z = N line thus moving
it away from the stability line into the region of too many neutrons and thus the
region of neutron decay.
Trend during  decay is
N  N - 2 and Z  Z - 2
ie along a line slope = 1,
parallel to N = Z
N
N=Z
Stability line
Z
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(ii) - decay occurs when a nuclide has too many neutrons for stability. This occurs
in the region above the stability line in an N versus Z graph. i.e. when N > Z. In
this case the neutron decays to a proton so Z goes to Z + 1
1
1
o
0
on  1p + -1e + 0  (proton stays in nucleus)
eg. 82Pb210  83Bi210 + -1eo + 00 
Trend during  decay is N
 N and Z  Z + 1 ie
along a line parallel to Z
axis
N
N=Z
Stability line
Z
(iii) + decay occurs when a nuclide has too many protons for stability. This
occurs in the region below the stability line when N is plotted against Z. i.e.
when N<Z. In this case the proton decays to a neutron decreasing Z by 1
1
1
o
1p  on + 1e +  (neutron stays in nucleus)
eg. 7N13  6C12 + 1eo +
Trend unstable
nuclide, too many protons,
N unchanged, Z  Z - 1
N
N=Z
Stability line
Z
2.
It often happens that the products of , - or + decay are left in excited states. If
the daughter nucleus is left in an excited state it will immediately undergo a
transition to a lower energy. This behaviour is similar to that of an atomic
system except that the energies involved are much higher. When the nucleus
falls from a state of higher to a state of lower energy  ray photon rather than a
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visible photon is emitted. It is in this way that the  rays emitted by radioactive
substances arise. This also explains why the  rays have discrete energies.
The following diagram illustrates the decay of 83 Bi212 (Bismuth) to thallium 81Tl208.
Six different energy particles are emitted accompanied by five different energy 
ray photons. (no photon is emitted for as it leaves thallium in the ground state).
The energies shown are measured relative to the ground state of thallium
6
5
4
3
2
1
0.617 MeV
0.492 MeV
0.472 MeV
0.328 MeV
 ray emitted
0.328 MeV
0.040 MeV
E1 = 6.086 MeV
0 MeV
The reaction is 83Bi212  81Tl208 + 2He4
(If you are astute you may note that some energy is missing 6.086 - 5.481 = .605
not .617 MeV. This is because the thallium nucleus also takes off some energy.)
3.
Naturally occurring radioactivity is essentially a result of heavy nuclei which have
Z > 82. e.g. U238 and U235. These decay by  decay into daughter nuclei.
92
92
Some of these heavy nuclei have a long half-life and hence are still present on the
Earth's surface.
The daughter nuclei are unstable and also decay by either  decay or - decay
because they have an excess of neutrons. This process continues until a stable
nuclide results. Consequently the naturally occurring radiations are , - and .
(the  accompany both the  and - modes of decay.)
4.
(a) (i) 6C12  7N14 + -1eo + o  o
(ii) 88Ra226  86Rn222 + 2He4
( Rn is Radon )
13
(iii) 7N  6C13 + 1eo + oo
(iv)13Al29  14Si29 + - + o  o
(v) 13Al24  12Mg24 + + + oo
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(b) aluminium has one stable isotope, (13Al27).13Al29 has an excess of neutrons and
so decays by neutron decay ( - decay) whereas 13Al24 has too many protons for
the 11 neutrons in it. The 11 neutrons are unable to compensate for the Coulomb
repulsive force on the charged protons. Consequently the protons decay to
neutrons ( + decay).
(c)  radiation. Some nuclei will be left in an excited state. The excess energy will
be lost as radiation when the nuclei drop to their more stable ground state.
5.
(a) on1  1p1 +
(b) 1p1 
on
1
o
-1e
o
1e
+
o
o
+ ono
6.
40
 20Ca40 + -1eo + o  o
19K
192
 76Os188 + 2He4
78Pt
210
 81Tl206 2He4
83Bi
210
83Bi
24
13Al


o
210
o
84Po
-1e + o 
24
+ + + ono
12Mg
7
(a)
(i)
(ii)
N 13  6 C 13 + +1 e 0 + 0  0
210
 83 Bi 210 + -1 e 0 + 0  0
82 Pb
7
(b) In any  decay reaction the mass of the products is less than the mass of the
parent nucleus. Some of the parent mass has been converted into energy. The
mass defect and hence the total released energy is constant and unique for a
given reaction. E max is a measure of this energy and so is also constant (note that
the daughter nucleus carries off an insignificant proportion of the energy
released as it is so massive compared to an electron.)
(c) The continuous spectrum of  particle energies indicates that most the 
particles are emitted with less energy than Emax. Experiments show that they
have not lost the energy in collisions with other atoms on the way out. This
means that if mass-energy is to be conserved the missing energy had to be
carried off by another particle which does not show up readily in experiments.
To be so "invisible" this particle must have
zero rest mass
no charge
very little interaction with matter
The particle is called the neutrino.
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8.
From conservation of momentum
Pbefore = P after

0 = P + PD
0 = m v + mD vD
v
mD


vD
m
v
mD
In size, then,
=
vD
m
Hence
(0.5)m v 2
E
=
ED
(0.5)mD v D 2
m  v 
 
=
mD  v D 
m
=
mD

 mD

 m



2
2
E
mD
=
ED
m
(b) From conservation of energy
E + ED = Q
but ED =
mD
= Q
m

E + E

 mD 
= Q
E  1
 m 
 mD  m 
E 
= Q
 mD 

(c)
 mD 
 Q
E = 
 mD  m 
 mD 
 Q
E = 
 mD  m 
208
=
Q
208  4
208
=
Q
212
= 98.1 % Q
mD
E
m
from (a)
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9.
Measurements of the energy spectrum of  and  rays show that they have "line
spectra", i.e. a few discrete values of  and  ray energies are associated with
each decaying nuclide. Analysis of the energies involved (taking into account
the fact that the  particle does not carry all of the energy released in a reaction)
suggests that the daughter nucleus may be formed in a stable or one or more
excited states. If left in an excited state excess energy is released by emission of
 ray photons as the nucleons in the daughter nuclei rearrange themselves into
the most stable ground state.
The following energy level diagram is based on measurement of the radiations
emitted in the decay of Bi212 to Thallium208
Bi ground state
6.20
E = 5.71 MeV
0.49
0.47
E = 6.20 MeV
Tl excited states
Tl ground state
0.33
MeV
above
ground
state
0.04
0
An  particle with energy 98 %
of each of these values is released. rays corresponding to each of these transitions
are observed
This strong inter-relation of the energy of the emitted  and  ray particles provides
the evidence for the existence of energy levels.