course index

Recall from last lecture:

29.1 Some Properties of Nuclei

Nuclear Stability

Nuclei are held together by the nuclear force, a force that is very strong over short distances -- strong enough to keep the protons of a nucleus bound together despite their Coulomb repulsion.

Not all nuclei are stable. Stability means that a nucleus will remain as it is, essentially forever. Nuclei that are not stable undergo nuclear decay. As a general rule, stable nuclei have as many, or more, neutrons as protons. And, the more protons a nucleus has, the more neutrons needed for stability.

29.2 Binding Energy

The mas of a nucleus is always less than the sum of the masses of the protons and neutrons that make up the nucleus. The difference in mass is due to the binding energy.

Ebind = (Zmp + Nmn - Mnucleus)c2
To get the binding energy per nucleon, divide by the mass number, A.

If we plot the binding energy per nucleon versus the number of nucleons, we see that there is a maximum binding energy for mass numbers near 60 (Figure 29.4). These are the most stable nuclei.

Example: P29.8

Compare the average binding energy per nucleon of 2412Mg and 8537Rb.

In appendix B, we find the atomic masses:
M(2412Mg) = (23.985042)u    and
M(8537Rb) = (84.911793)u.
From table 29.1 we find the mass of the proton and neutron are:
mp = (1.007276)u    and
mn = (1.008665)u.
2412Mg has 12 protons and 12 neutrons, so the binding energy is:
Ebind = 12(1.007276)u + 12(1.008665)u - (23.985042)u = 0.206u = 0.206(931 MeV/c²) = 192 MeV/c²
The number of nucleons is 24, so the binding energy per nucleon is:
Ebind / A = 8.00 MeV/c²

8537Rb has 37 protons and 48 neutrons, so:
Ebind = 37(1.007276)u + 48(1.008665)u - (84.911793)u = 0.773u = 0.773(931 MeV/c²) = 720 MeV/c²
The number of nucleons is 85, so the binding energy per nucleon is:
Ebind / A = 8.47 MeV/c²

29.3 Radioactivity

Radioactivity is the spontaneous emission of radiation by an unstable nucleus. There are three types of radiation which are emitted by various nuclei:

  1. alpha (a) rays (particles) which are 4He nuclei (2 protons and 2 neutrons),
  2. beta (b) rays (particles) which are electrons, or, in a few cases, positrons (the anti-particle of the electron), and
  3. gamma (g) rays which are high energy photons.

These three types of radiation can be distinguished by their behaviors. Gamma rays are not deflected (bent) by a magnetic field; alpha particles are bent as for a positive charge, and most beta particles are bent as fro a negative charge.

They are also distinguished by their penetrating power. Alpha particles are barely able to penetrate through a sheet of paper. Beta particles can penetrate a few millimeters of aluminum or other light material. Gamma rays can penetrate a few centimeters of lead or other heavy material.

The Decay Constant and Half-Life

Radioactive decay is a stochastic (random, probabilistic) process. It does seem to be a universal feature of radioactive decay that the probability for a radioactive nuclei to decay during a period of time is a constant for that type of nuclei. If we have N radioactive nuclei, then during time Dt, the number, DN, that decay is given by

DN = l N Dt
where l is the decay constant for that type of nuclei.

The activity, R, of a sample is the number of decays per second

R = DN / Dt = l N

This leads to the result that, starting initially with N0 radioactive nuclei, the number, N, remaining (not having decayed) a time t later is

N = N0 e-lt
where e = 2.718... is the base of the natural logarithms. This is an exponential decay function.

The time it takes for half the initial nuclei to decay is called the half-life, T1/2, where the half-life is related to the decay constant by:

T1/2 = ln(2) / l = 0.693 / l

There are 2 units for activity. The first is the curie (Ci), defined as 1Ci = 3.7×1010 decays/s. The second is the SI unit, the becquerel (Bq), defined as 1Bq = 1 decay/s.

Example: P29.14

A drug tagged with 4399Tc (half-life = 6.05h) is prepared for a patient. If the original activity of the sample was 1.1×104Bq, what is its activity after it has sat on the shelf for 2.0h?

The activity is proportional to the number of undecayed nuclei in the sample, R = lN. If the inital activity is R0 = lN0 = 1.1×104Bq, then after time t, the activity will be
R = lN = lN0 e-lt = R0e-lt.
We can determine l from the half-life:
l = 0.693 / T1/2 = 0.693 / 6.05h = 0.1145/h.
Therefore:
R = (1.1×104Bq)e-[(0.1145/h)(2.0h)] = 0.87×104Bq = 8.7×103Bq.

© Robert Harr 2000