Lesson

Alpha decay

In our every day language the word decay is often taken to mean rotting.  This is not the case here.  If a radioactive substance changes into another substance because particles are emitted from its nucleus, we say that the original parent element is decaying.

In the case of alpha decay, an α particle, or ,  is emitted from the parent nucleus.

Suppose the parent substance is .  Actually if you look up Z=88 in the periodic table, you will see that X is radium.  So, the parent element is .

question

If an alpha particle is emitted from the parent element, what will be the A and Z numbers of the daughter element that is left behind?

answer

226 will diminish by 4 and 88 will diminish by 2.  Therefore, in the daughter element, A=222 and Z=86. 

question

Look up the periodic table to determine which element has A=222 and Z=86.

answer

Radon, or

question

Write an equation to show that the total number of nucleons is conserved.

answer

See below.  When the alpha particle leaves the radium nucleus, it takes with it 2 protons and 2 neutrons for a total of A = 4 nucleons.  This means that the radium decays to a daughter element with 2 fewer neutrons and 2 fewer protons (4 nucleons).  But, when the nucleons are added up on the right-hand side, they equal the original number of nucleons on the left-hand side.

The colourful notation above is for a specific situation caused by the decay of radium into radon.  You should be able to see that this notation of a special case suggests a more general expression for all alpha decay.  For instance, let the parent element be .  Then the daughter element can be represented by   The 4 is subtracted from A, and the 2 is subtracted from Z because an alpha particle is ejected from .  The general notation for alpha decay is therefore

 

The demonstration below illustrates alpha decay.  Because the animation automatically moves from one step to another after you have started it, you may need to play it 5 or 10 times.  For the first few times just look at the picture.  Then, in the next few runs, read the accompanying text to the left of the picture.  Play it until it makes sense.

Now let's do a few exercises...

Practice exercise 1

An unstable polonium atom spontaneously emits an alpha particle and transmutes into a daughter element (Y).  Determine the A and Z numbers of the new element and refer to the periodic table to identify the new element.  Write an equation to show the nuclear process.

Solution

Since the α particle is , the atomic mass number for the daughter element will be A = 218 - 4 = 214;   and the atomic number will be

Z = 84 - 2 = 82.

Therefore the nuclear notation for the daughter element can be written as 

According to the periodic table   The unstable polonium decayed into lead.

The nuclear equation is

Beta decay

In Lesson 02 of this section you learned that beta decay was of two types, β- being an electron, and β+ being a positron.  A positron is an anti-electron in the sense that it is very much like an electron, except its charge is different. 

[Care is needed here because particles and anti-particles have more differences that just opposite charge.  For example, there are neutrons and anti-neutrons, and, as you know, neutrons have zero charge.  In a complicated way, antimatter is sort of the mirror image of matter, but all of that is too difficult to tackle in high school.]

Anyway, back to beta decay.  First, let us deal with β- , an emitted electron.  You must realize that such an electron is not an electron that was orbiting the nucleus.  Rather, it is an electron that is created in the nucleus.  The theory is that the electron is produced when a neutron decays to a proton (really weird?!)  Look at the story in the picture.

The picture shows a neutron changing into a proton, emitting an electron as it does so.  Note that the amount of charge is conserved.  It was zero before the decay, and adds up to zero after the decay.  The researchers, however, discovered a problem.  The emitted electron did not have as much energy as it should.  It was concluded that there must be another particle emitted along with the electron to account for the missing energy.  Here is where things become really weird.  The missing particle was said to have zero mass (or very nearly zero) and zero charge!  It was given the name neutrino (little neutron?).  Also, for reasons that we can't get into here, it was necessary that the particle accompanying the β- decay be an antineutrino and not a neutrino.  The symbol for neutrino is and the symbol for antineutrino is . 

So, the next picture tells the complete story of β- decay.

Now for the touchy part: writing an equation to show what happens when a parent element decays to a daughter element by the process of β- decay.

As usual, let the parent element be One of the neutrons in this nucleus is going to emit an electron and decay into a proton.  This means there will be one more proton in daughter Y than in parent X.  There will be no change in A because a neutron changing into a proton does not affect A.  (Remember A = # of protons + # of neutrons).  So daughter Y will look like this:   Don't forget the other two particles: the electron and the antineutrino.  Since the electron and the antineutrino are so light compared to the daughter nucleus, the atomic mass number (A) is zero for both.  For the antineutrino, Z is also zero because it has no charge.  However, for the emitted electron which has one unit of negative charge, Z = -1.  All of this goes together to give the following notation for β- decay:

If you like, you can simplify the β- decay equation by omitting the antineutrino term.  The notation then becomes (with a slight rearranging):

Can you see that the number of particles and the amount of charge is conserved?  The number of particles (protons + neutrons)  is A on the left-hand side, and A +0 = A on the right-hand side.  The amount of charge is Z on the left-hand side and Z + 1 + (-1) = Z on the right-hand side.

Practice exercise 2

Carbon-14 has 6 protons (and therefore 8 neutrons) in its nucleus.  One of the neutrons suffers β- decay and the carbon transmutes to a daughter nucleus.  Write out this reaction in symbols and determine the name of the daughter element.

Solution

The symbol for the parent element is   β- decay means that one of the 8 neutrons will emit an electron and change into a proton.  Therefore the Z number of the daughter will be one more than the Z number of the parent.  Recall that the general equation for β- decay is

      (Here the antineutrino is omitted for the sake of simplicity).

For the expression becomes

The daughter element is therefore   which, according to the periodic table, is   The carbon-14 has transmuted into

nitrogen-14.

Again, notice that in β- decay the A number does not change.  For carbon-14, 14 = 6 protons + 8 neutrons; and, for nitrogen-14,

14 = 7 protons + 7 neutrons because one of the neutrons transmuted into a proton when the electron was emitted.

We've beat the life out of β- decay.  Let's go on to β+ decay.  (We know you can't wait!)

As the sign suggests, β+ decay is somewhat opposite to what goes on in β- decay.  Recall that in β- decay an electron was emitted as a neutron decayed into a proton.  In β+ decay a positron (e+) is emitted as a proton decays into a neutron.  Also, playing the earlier role of the antineutrino, we now have a neutrino accompanying the positron in order that energy be conserved.

In β- decay the Z number of the daughter element increased by 1.  According to the last paragraph, in β+ decay the Z number of the daughter element will decrease by 1.  The decrease occurs because a charged proton changes into an uncharged neutron.

Summing up the last two paragraphs, the nuclear equation for a parent element transmuting into daughter Y by β+ decay is

Again, to keep things simple, the expression is sometimes written without the neutrino:

Practice exercise 3

Sodium-22 has 11 protons (and therefore 11 neutrons) in its nucleus.  One of the protons suffers β+ decay and the sodium transmutes to a daughter nucleus.  Write out this reaction in symbols and determine the name of the daughter element.

Solution

(You should scroll back to practice exercise 2 and identify the differences between  it and this exercise.)

The symbol for the parent element is   β+ decay means that one of the 11 protons will emit a positron and change into a neutron.  Therefore the Z number of the daughter will be one less than the Z number of the parent.  Recall that the general equation for β+ decay is

      (Here the antineutrino is omitted for the sake of simplicity).

For the expression becomes

The daughter element is therefore   which, according to the periodic table, is   The sodium-22 atom has transmuted into

neon-22.

Again, notice that in β+ decay the A number does not change.  For sodium-22, 22 = 11 protons + 11 neutrons; and, for neon-22,

22 = 10 protons + 12 neutrons because one of the protons transmuted into a neutron when the positron was emitted.

The next demonstration shows beta decay.  Is it β-  or β+ ?   How do you know?   Because the animation automatically moves from one step to another after you have started it, you may need to play it 5 or 6 times.  For the first few times just look at the picture.  Then, in the next few runs, read the accompanying text to the left of the picture.  Play it until it makes sense.

Gamma radiation

In earlier lessons you learned that when an electron falls from a high energy level to a low energy level, a photon of electromagnetic energy is emitted.  Even though a gamma ray is a photon of electromagnetic energy it is not produced in that way. 

question

So what's so special about the way that a γ ray is produced?

answer

A γ ray is produced not because an excited electron falls to a lower level, but because an excited nucleus decays to a lower level.

question

I can sort of understand how an electron can be in an excited state (by being in a higher orbit), but, how can a nucleus be in an excited state?

answer

One way for it to be excited is if it is a daughter nucleus that has just been transmuted from a parent nucleus (birth is pretty exciting!).

You have already done three ways that a daughter element can be produced: α decay, β- decay, and β+ decay.  In these processes the daughter element was always on the right-hand side of the nuclear equation.  If the daughter element is in an excited state, then a γ ray may also have to added to the right-hand side of all three equations.  Since a γ ray has no mass and no charge, it is just a matter of tacking it on.  You don't have to worry about changing the A and Z numbers.  For example, a complete β- decay process including the emission of a gamma ray looks like this:

Brace yourself--your textbook (p. 749) describes another way to produce an excited daughter element.  The process is called electron capture. As the phrase suggests, in electron capture the parent nucleus absorbs an electron.  This means one of the protons decays into a neutron.  (You can picture the electron "canceling" the positive charge of the proton).  Because the parent nucleus absorbs an electron, for the first time you see two terms on the left-hand side of the nuclear equation.   A general equation is given below:

Just concentrate on the right-hand side of the last colourful equation.  Realizing that the γ ray comes from the daughter nucleus, and omitting reference to the neutrino, a tidier expression is

Because the γ ray has no mass and no charge, there is no effect on the A and Z numbers.  (But we said that before).

Practice exercise 4

Determine the value of x and y in each nuclear equation where P = parent and D = daughter.  Name the type of decay.

Solution

a)  An electron is emitted from the parent nucleus.  That's β- decay.  This means a neutron has decayed into a proton , which means

x = 82 + 1 = 83.

b)  This is α decay because 2 neutrons and 2 protons are emitted as a single particle.  x = 84 - 2 = 82.        y = 210 - 4 = 206.

c)  A, the unified atomic mass, did not change, but the number of protons or positive charges increased by 1.  Therefore, a neutron has decayed into a proton.  This means that an electron was emitted from the parent nucleus.  This is  β- decay. 

d)  This is gamma radiation.  With the emission of a gamma ray the parent nucleus does not change.  That is, the atomic number and atomic mass number stay the same.   x = 88 and y = 226.

e)  The tells us that this is β- decay.   214 = x + 0;  x = 214.

83 = y + (-1);  y = 84.

f)  The tells us that this is α decay.  x = 224 + 4 = 226;

y = 86 + 2 = 88.

g)  A decreases by 4 and Z decreases by 2.  Therefore the emitted particle (x) must be . 

h)  This is gamma radiation.  Because a gamma ray has no mass and no charge, the A and Z numbers of the parent nucleus do not change.  x  =  .

Energy released during radioactive decay

In Lesson 1 of this section you were introduced to "mass defect" and mass difference" when two atomic particles combined.  In this lesson we have dealt with atomic particles that "fly apart".  Once again there a mass defect.  It turns out that the combined mass of the daughter products is less than the mass of the parent element before decaying occurred.

We will consider only α decay in detail. 

Let mP = the mass of the parent element

Let mD = the mass of the daughter element

Let mα  = the mass of the alpha particle emitted from the parent

The classical law of conservation of mass states that the mass of the parent element should equal the sum of the other two masses.  That is,

mP =  mD +  mα .

This statement is not true for nuclear reactions.  For nuclear decay, the mass of the daughter elements summed together is less than the mass of the parent:

(mD +  mα )   <   mP

question

Where does the mass go?

answer

Into energy, according to E = Δmc2 .   Here Δm = mP - (mD +  mα).

Recall that we said at the beginning of this section that only alpha decay would be considered in detail.  In fact, beta and gamma decay are even more straightforward.  This is because no mass (or nearly zero mass) is associated with these two radiations.  Therefore,

Δm = mP - mD . 

Practice exercise 5 on page "f" uses the concept of mass defect and mass difference.

Item 5 in the self-test is based on beta decay.

Mass defect and binding energy

As the term suggests, "binding energy" is the energy that must be overcome in order to break a parent nucleus into its daughter components.  This binding energy is directly related to the "mass difference" (∆m) which occurs when the nuclear reaction takes place.  It is the difference in mass which provides the binding energy and this energy can be described by Einstein's formula, E = mc2.  Certain isotopes have a very small mass defect when they decay--hence the binding energy of such isotopes is also small.  When the binding energy of an isotope is small, the nucleus may break up (decay) spontaneously.  Such isotopes are said to be radioactive.

Practice exercise 5

Element P (the parent element) has a mass of 237.025 567 u.

Element D (the daughter element) has a mass of 233.016425 u.

The emitted α particle has a mass of 4.002602 u.

How much energy is available as kinetic energy for the daughter nucleus and the α particle after transmutation?

Solution

Step 1--read and understand the story of the problem

The most important thing to remember is that the classical idea of conservation of mass no longer applies.  Still, mass is not really lost.  Rather, in nuclear reactions, it is transformed into energy . 

Step 2--list the givens and the unknown

mP = 237.025 567 u

mD = 233.016425 u

mα = 4.002602 u

1 u = 931.5 Mev/c2

Δm = mass difference = ?

E     = ?

Step 3--plan a strategy

Find Δm from Δm = mP - (mD +  mα). 

Convert mass units by using  1 u = 931.5 Mev/c2.  (You may wish to review this conversion factor in  Unit 03  Section 06  Lesson 01.)

Find the energy released by using  E = Δmc2 . 

Step 4--carry out the plan

Δm = mP - (mD +  mα)

      =  237.025 567 u   -  ( 233.016425 u  +  4.002602 u)

      =  0.00654 u

Converting 0.00654 u in Mev/c2  :

Δm = 0.00654  x  931.5 Mev/c2  =  6.09 MeV/c2

Therefore,

E = Δmc2  =  6.09 MeV/c2   x  c2  

E  =  6.09 Mev.

Step 5--communicate

The energy available to provide kinetic energy to the products of the decay is 6.09 Mev.

Activity

Continue with the major assignment of Unit 03  Section 01  Lesson 01.

1.  Write the reaction equations for:

  1. the β- decay of
  2. the α decay of
  3. the β+ decay of
  4. electron capture for the isotope

2.  Fill the missing particle or nucleus:

In your textbook:

on p. 751--do #1-#6.

on p. 772--do #7-#10;  #12-#13.  You may want to discuss #7-#10 with your teacher.  In #12 think about conservation of mass.

on p. 774--do #43-#46, #48, #51 

For #48 and #51 use 1 u = 931.5 MeV/c2  and E = Δmc2 .  You can review mass difference in Unit 03  Section 06  Lesson 01.

Answers

When answering #1 and #2 make sure that the "Z" and "A" numbers are conserved.  Once you know the "A" number of the daughter nucleus, find the element in the periodic table (Appendix C of your textbook).

1 a)

  b) 

  c) 

  d) 

2  a)         b)          c)          d)         e) 

Test Yourself

1 u  =  931.5 MeV/c2