BMAT Physics Notes: Radioactivity


Structure of an Atom

There is a very small central nucleus within which the protons and neutrons exist.
Around the nucleus the electrons move in circular orbits.
Most of the mass of an atom is in the nucleus.
Neutrons have a mass ≈ mass of proton.
Electrons have a much smaller mass than either a proton or neutron.
Neutrons have zero charge.
​For an (electrical) neutral atom,  number of protons = number of electrons.
If an atom has gained (or lost) one or more electrons it is called an ion.
The electrons being negatively charged experience an attractive force from the positively charged protons in the nucleus. It is this force that keeps the electrons moving in circular paths. 

​Some definitions
 
Nucleon: This is a particle inside the nucleus (i.e a proton or a neutron).
 
Atomic mass number A ( or nucleon number): This is the number of nucleons (neutrons and protons) in the nucleus.
 
Atomic number Z: This is the number of protons in the nucleus.
Nuclide: Any neucleus.
 
Isotope: These are nuclei  with the same atomic number (i.e. protons) but with different mass number (i.e. different number of neutrons).
​​
Mass Number A
(or Nucleon Number)
The number of nucleons
​(protons + neutrons)
in the nucleus.                                                                                               
Atomic Number Z
(or Proton number)
The number of protons        
in the nucleus. This number
determines what element it is.

Example 1
​
How many neutrons in an atom of gold?   __________
​

​Radioactivity
 
Some elements have unstable nuclei. They are ‘breaking up’ or decaying. Such elements are said to be radioactive.
It is not possible to predict which nucleus or when a nucleus will decay. It is a random process.
 
When a nucleus decays three types of emission occur (sometimes called nuclear radiation).
When alpha and beta decay occurs a new chemical element is formed.
​

​Basic Properties

ALPHA

BETA
GAMMA​

Radioactive Decay Equations
 
[NOTE: sometimes in decay processes the initial nucleus is called the parent nuclide and the final nucleus is called the daughter nuclide]
​

​ALPHA (α) DECAY

​This can be represented as follows:
​An example:
Note:
  • The atomic mass number A is conserved.
  • The atomic number Z is conserved.
    ​​

BETA DECAY.  (β-)
​
We have seen that a b- particle is an electron. However, there are no electrons inside the nucleus. So where do they come from?
 
It was discovered that under certain conditions neutrons in the nucleus are unstable and decay into a proton and an electron. The proton stays in the nucleus but the electron is ejected from the nucleus as a beta (β -) particle.
​An example:

GAMMA (γ) DECAY
 
Since gamma rays are electromagnetic waves they therefore have no mass or charge. This means the atomic mass number A and atomic number Z do not change in g-decay.
 
The nucleus of any nuclide can have ‘excited’ states like energy levels of electrons. An excited nucleus may return to a more stable state by losing energy in the form of a g-ray.
​An example:  

​Example 2

​Deflection in an Electric Field
                                
Suppose 2 parallel plates are charged, one positive (+) and the other negative (-) as shown.
 
This sets up an electric field between the plates (not shown).
 
A beam consisting of a, b and g rays enters the field.
The γ rays pass straight through with no deflection. This is because they are not charged and the electric field has no effect on them.                                  
 
The α particles are attracted to the bottom negative plate as shown. This is because a particles are positively charged and opposite charges attract.
 
The β particles are attracted to the top positive plate as shown. This is because the b particles are negatively charged.
 
The β particles are deflected much more than the a particles because b particles are much lighter than a particles.

Deflection in a Magnetic Field
 
The γ rays are undeflected.
 
The α and β rays follow circular paths whilst IN the field but in opposite directions.
The direction of the deflecting force is given by the left-hand rule.
 

​Half-Life (T)

The half-life of a radioactive substance is the time taken for the number of undecayed nuclei  to halve.
Suppose a radioactive element has a half-life of 10 minutes.
Let there be N undecayed nuclei at time t = 0.

After 10 minutes ( 1 half-life) ½ the nuclei will have decayed, so number of undecayed nuclei  = N/2
After 20 minutes ( 2 half-lives) ½ of the remaining nuclei will have decayed, so number of undecayed nuclei  =  ½ (N/2) = N/4
After 30 minutes ( 3 half-lives) ½ of the remaining nuclei will have decayed, so number of undecayed nuclei  =  ½ (N/4) = N/8 and so on.
​
Since the number of atoms or mass is proportional to the number of nuclei many questions refer to the mass in grams.
 
Example:  A radioactive element has a half-life of 1 hour and an initial mass of 2g.
 
Calculate the undecayed mass after 4 hours.
 
After 1 h undecayed mass = ½ (2) = 1g
After 2 h undecayed mass = ½ (1) = 0.5g
After 3h undecayed mass = ½ (0.5) = 0.25g
After 4h undecayed mass = ½ (0.25) =0.125g
​

​Background Radiation
 
There are radioactive materials all around us. Some of this is ‘man-made’ but most of it is produced naturally in the Earth.
This is called the background radiation and has been around for millions of years.
 
Sometimes the background count due to this is measured.
If the background count is given then this has to be subtracted from the total count rate when calculating half-lives.
​
​Example 3

The background count in a room is 40 counts per minute.
A detector is placed in front of a radioactive source in the room. Initially the count rate is 1080 per minute.
What is the count rate after two half-lives of the source?

A   270 counts per minute         B   250 counts per minute         C   260 counts per minute          D   240  counts per minute
 

​Graphical Method
 
Sometimes the half-life of an element is calculated from a suitable graph as shown.
 
To determine the half-life:
 
(i) Choose a convenient point on the graph near the start e.g. 4.0g and corresponding time t = 0.1 mins
(ii) read off the time taken for the mass to halve e.g. 2.0 g, this occurs at about time t = 2.4 mins
(iii) time interval is (2.4 – 0.1) mins
      = 2.3 mins = half-life.
​
​       
​       Mass (g)
              time (mins)               
 
Example Answers

​Example 1:   118 neutrons

​Example 2:   Answer C

Example 3:   Answer C