Physics 6C
Nuclear Physics and
Radioactivity
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Mass Defect (Binding Energy)
A stable nuclei is comprised of Z protons and N neutrons, bound together by the strong nuclear
force. The fully assembled nucleus is less massive than the individual protons and neutrons, if
they are weighed separately. The difference between these masses is called the mass defect.
From Einstein’s E=mc2 equation, we can also consider this mass defect to be binding energy.
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Mass Defect (Binding Energy)
A stable nuclei is comprised of Z protons and N neutrons, bound together by the strong nuclear
force. The fully assembled nucleus is less massive than the individual protons and neutrons, if
they are weighed separately. The difference between these masses is called the mass defect.
From Einstein’s E=mc2 equation, we can also consider this mass defect to be binding energy.
As an example, we will calculate the mass defect for oxygen-16, the most common isotope of
oxygen. This nucleus has 8 protons and 8 neutrons.
We can look up the atomic mass of O-16 in a table (below).
This is the mass of a neutral atom of
O-16. We will need to account for the
8 electrons in our calculation.
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Mass Defect (Binding Energy)
Now that we have the mass of the assembled atom (mO16=15.994915u), we need to add up
the mass of all the parts. Also listed in a table are the masses we need.
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Mass Defect (Binding Energy)
Now that we have the mass of the assembled atom (mO16=15.994915u), we need to add up
the mass of all the parts. Also listed in a table are the masses we need.
We have 8 protons, 8 electrons, and 8
neutrons. As a shortcut we can group
the protons and electrons together and
use 8 hydrogen atoms instead. Here is
the calculation:
8mH  8mn  8(1.007825u)  8(1.008665u)
mtotal  16.13192u
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Mass Defect (Binding Energy)
Now that we have the mass of the assembled atom (mO16=15.994915u), we need to add up
the mass of all the parts. Also listed in a table are the masses we need.
We have 8 protons, 8 electrons, and 8
neutrons. As a shortcut we can group
the protons and electrons together and
use 8 hydrogen atoms instead. Here is
the calculation:
8mH  8mn  8(1.007825u)  8(1.008665u)
mtotal  16.13192u
Now we just subtract to find the defect:
m  16.13192u  15.994915u
m  0.137005u
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Mass Defect (Binding Energy)
Now that we have the mass of the assembled atom (mO16=15.994915u), we need to add up
the mass of all the parts. Also listed in a table are the masses we need.
We have 8 protons, 8 electrons, and 8
neutrons. As a shortcut we can group
the protons and electrons together and
use 8 hydrogen atoms instead. Here is
the calculation:
8mH  8mn  8(1.007825u)  8(1.008665u)
mtotal  16.13192u
Now we just subtract to find the defect:
m  16.13192u  15.994915u
m  0.137005u
This value is in atomic mass units. We can also convert to kg or energy, either in
Joules or MeV. The conversion factors are also in the book.
m  (0.137005u)(931.5 MeV
)
u
m  127.6MeV
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Nuclear Stability
Oxygen-16
is here.
This chart shows the binding energy per nucleon for all the stable nuclei. The
higher the value, the more stable the nucleus will be.
Our calculation for oxygen-16 gives 7.98 MeV/nucleon. The most stable nuclei are
metals like iron, cobalt and nickel, with values just below 9 MeV/nucleon.
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Types of Radioactive Decay
Alpha Decay (α)
When a nucleus is too large to be stable, it will often emit an alpha-particle, which is really
just a helium nucleus (2 protons and two neutrons). The alpha particle carries away some
energy, and the resulting nucleus has 2 fewer protons and 2 fewer neutrons (and thus is 2
places lower on the periodic table).
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Types of Radioactive Decay
Alpha Decay (α)
When a nucleus is too large to be stable, it will often emit an alpha-particle, which is really
just a helium nucleus (2 protons and two neutrons). The alpha particle carries away some
energy, and the resulting nucleus has 2 fewer protons and 2 fewer neutrons (and thus is 2
places lower on the periodic table).
Beta Decay (β)
When a nucleus has too many excess neutrons, it may undergo beta-minus decay,
whereby a neutron decays into a proton and an electron. The electron is emitted, leaving
the nucleus with one more proton than before (and one less neutron). So the result is an
atom that is one step higher on the periodic table.
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Types of Radioactive Decay
Alpha Decay (α)
When a nucleus is too large to be stable, it will often emit an alpha-particle, which is really
just a helium nucleus (2 protons and two neutrons). The alpha particle carries away some
energy, and the resulting nucleus has 2 fewer protons and 2 fewer neutrons (and thus is 2
places lower on the periodic table).
Beta Decay (β)
When a nucleus has too many excess neutrons, it may undergo beta-minus decay,
whereby a neutron decays into a proton and an electron. The electron is emitted, leaving
the nucleus with one more proton than before (and one less neutron). So the result is an
atom that is one step higher on the periodic table.
Gamma Decay (γ)
Sometimes another decay process will leave the resulting nucleus in an excited state, when
the nucleus drops down a gamma-ray photon is emitted to carry away the excess energy.
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Decay Rates
For any decay process there will be an associated HALF-LIFE, which describes how long it
should take for half of the radioactive nuclei in a sample to undergo the decay.
We can keep track of this in several different ways, but it all boils down to the same basic
formula. Here is an exponential decay formula:
N(t)  N0  et
This will count the remaining radioactive nuclei in a sample at any given time. You can
also use it for the activity of a sample.
N0 is the starting amount, and λ is the “decay constant” (related to the half-life).
T1
2
ln( 2)


The fundamental relationship you need is for the decay rate (this is also called ‘ACTIVITY’:
N
   N
t
Units for activity are usually decays/second.
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Example: Suppose a radioactive sample has an activity of 100,000 decays per second.
When the same sample is measured again 10 days later, the activity has dropped to
80,000 decays per second. What is the half-life of the sample?
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Example: Suppose a radioactive sample has an activity of 100,000 decays per second.
When the same sample is measured again 10 days later, the activity has dropped to
80,000 decays per second. What is the half-life of the sample?
We can use our basic exponential decay equation and solve for the decay constant:
N(t)  N0  e t
80,000  100,000  e (10days)
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Example: Suppose a radioactive sample has an activity of 100,000 decays per second.
When the same sample is measured again 10 days later, the activity has dropped to
80,000 decays per second. What is the half-life of the sample?
We can use our basic exponential decay equation and solve for the decay constant:
N(t)  N0  e t
80,000  100,000  e (10days)
0.8  e (10days)
ln(0.8)  (10days)
1
  0.0223 day
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Example: Suppose a radioactive sample has an activity of 100,000 decays per second.
When the same sample is measured again 10 days later, the activity has dropped to
80,000 decays per second. What is the half-life of the sample?
We can use our basic exponential decay equation and solve for the decay constant:
N(t)  N0  e t
80,000  100,000  e (10days)
0.8  e (10days)
ln(0.8)  (10days)
1
  0.0223 day
Now just use our formula for half-life:
ln(2)
T12 
 31days
1
0.0223 day
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Example: Carbon-14 is a radioactive isotope with a half-life of 5730 years. It is
sometimes used to find the age of archaeological finds. Suppose a bone is dug up and it is
determined that 20% of the original amount of 14C remains. How old is the bone?
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Example: Carbon-14 is a radioactive isotope with a half-life of 5730 years. It is
sometimes used to find the age of archaeological finds. Suppose a bone is dug up and it is
determined that 20% of the original amount of 14C remains. How old is the bone?
We can use our basic exponential decay equation and solve for the time.
N(t)  N0  e t
20%  N0  N0  e t
0.2  e t
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Example: Carbon-14 is a radioactive isotope with a half-life of 5730 years. It is
sometimes used to find the age of archaeological finds. Suppose a bone is dug up and it is
determined that 20% of the original amount of 14C remains. How old is the bone?
We can use our basic exponential decay equation and solve for the time.
N(t)  N0  e t
20%  N0  N0  e t
0.2  e
 t
The decay constant is found
from the half-life equation.
ln(2)
ln(2)


T12
5730yr
  1.21  10 4
1
yr
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Example: Carbon-14 is a radioactive isotope with a half-life of 5730 years. It is
sometimes used to find the age of archaeological finds. Suppose a bone is dug up and it is
determined that 20% of the original amount of 14C remains. How old is the bone?
We can use our basic exponential decay equation and solve for the time.
N(t)  N0  e t
The decay constant is found
from the half-life equation.
20%  N0  N0  e t
0.2  e
 t
ln(0.2)  (1.21  10 4
t  13,300years
1
yr
)t
ln(2)
ln(2)


T12
5730yr
  1.21  10 4
1
yr
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Radiation Doses
When radiation is absorbed by living tissue we need a way to measure the effects.
There are two steps to the measurement.
1) How much energy was absorbed (per kg of tissue)?
The rad is the more commonly used unit:
1 rad = 0.01 J/kg
2) What type of radiation is it?
Different types of radiation affect living tissue in different ways. To account for this we
have a multiplicative factor called RBE (relative biological effectiveness).
Multiply together to get the “biologically equivalent dose”.
Equivalent dose (REM) = RBE X absorbed dose (RAD)
An example follows…
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Example: Alpha-particles with an RBE of 13 deliver a 32 mrad whole-body radiation dose
to a 69 kg patient.
a) What dosage, in rem, does the patient receive?
b) How much energy is absorbed by the patient?
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Example: Alpha-particles with an RBE of 13 deliver a 32 mrad whole-body radiation dose
to a 69 kg patient.
a) What dosage, in rem, does the patient receive?
b) How much energy is absorbed by the patient?
For part a) we multiply to get the equivalent dose in rems:
dose  (13)  (32  10 3rad)  0.416rem
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Example: Alpha-particles with an RBE of 13 deliver a 32 mrad whole-body radiation dose
to a 69 kg patient.
a) What dosage, in rem, does the patient receive?
b) How much energy is absorbed by the patient?
For part a) we multiply to get the equivalent dose in rems:
dose  (13)  (32  10 3rad)  0.416rem
For part b) just use the definition of rad:
J
1rad  0.01 kg
J )  3.2  10  4
32mrad  (32  10 3 )(0.01 kg
Eabsorbed  (69kg)(3.2  10  4
J)
kg
J
kg
 0.022 J
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