Radioactive disintegration/decay can be initiated in an industrial laboratory through two chemical methods:

a) nuclear **fission**

b) nuclear **fusion.**

**a)Nuclear fission**

Nuclear fission is the process which a fast moving neutron bombards /hits /knocks a heavy **unstable** nuclide releasing l**ighter** nuclide, **three** daughter neutrons and a large quantity of **energy. **

Nuclear fission is the basic chemistry behind **nuclear bombs** made in the nuclear reactors.

The three daughter neutrons becomes again fast moving neutron bombarding / hitting /knocking a heavy unstable nuclide releasing lighter nuclides, three more daughter neutrons each and a larger quantity of energy setting of a **chain reaction**

**Examples of nuclear equations showing nuclear fission**

^{1}_{0}^{ }n + ^{235} _{b} U -> ^{90}_{38}Sr + ^{c} _{54}Xe + 3^{1}_{0}n + a

^{1}_{0}^{ }n + ^{27}_{13}Al -> ^{28}_{13}Al + y + a

^{1}_{0}^{ }n + ^{28}_{a}Al -> ^{ b}_{11}Na + ^{4}_{2}^{ }He

^{ a}_{0 }n + ^{14}_{7}N -> ^{14}_{b}C + ^{ 1}_{1}H

^{1}_{0 }n + ^{1}_{1}H ^{ }_{ }^{ }-> ^{2}_{1}H ^{ }_{ } + a

^{1}_{0 }n + ^{235} _{92} U -> _{ }^{95} _{42} Mo + ^{ 139} _{57} La + 2^{1}_{0}n + 7 a

**b) Nuclear fusion**

Nuclear fusion is the process which **smaller **nuclides join together to form **larger **/ heavier nuclides and releasing a large quantity of **energy**.

Very high temperatures and pressure is required to overcome the repulsion between the atoms.

Nuclear fusion is the basic chemistry behind solar/sun radiation.

Two daughter atoms/nuclides of Hydrogen fuse/join to form Helium atom/nuclide on the surface of the sun releasing large quantity of energy in form of heat and light.

^{2}_{1}H + ^{2}_{1}H -> ^{a}_{b}He + ^{1}_{0}n

^{ 2}_{1}H + a -> ^{3}_{2}He

^{ 2}_{1}H + ^{2}_{1}H -> a + ^{1}_{1}H

4 ^{1}_{1}H -> ^{4}_{2}He + a

^{14}_{7}H + a -> ^{17}_{8}O + ^{1}_{1}H

**C: HALF LIFE PERIOD (t ^{1}/_{2})**

The half-life period is the **time** taken for a radioactive nuclide to spontaneously decay/ disintegrate to **half** its **original** mass/ amount.

It is usually denoted **t ^{1}/_{2}**.

The rate of radioactive nuclide disintegration/decay is **constant** for each nuclide.

**The table below shows the half-life period of some elements.**

Element/Nuclide | Half-life period(t ^{1}/_{2} ) |

^{238} _{92} U | 4.5 x 10^{9} years |

^{14} _{6} C | 5600 years |

^{229} _{88} Ra | 1620 years |

^{35} _{15} P | 14 days |

^{210} _{84} Po | 0.0002 seconds |

The **less** the half life the** more unstable** the nuclide /element.

The half-life period is determined by using a Geiger-Muller counter (**GM tube**)

.A GM tube is connected to ratemeter that records the** count-rates per unit time**.

This is the rate of decay/ disintegration of the nuclide.

If the count-rates per unit time **fall **by **half**, then the **time** taken for this **fall** is the half-life period.

**Examples**

**a)A radioactive substance gave a count of 240 counts per minute but after 6 hours the count rate were 30 counts per minute. Calculate the half-life period of the substance.**

If t ^{1}/_{2 } = x

then 240 –x–>120 –x–>60 –x—>30

From 240 to 30 =3x =6 hours

=>x = t ^{1}/_{2} = ( 6 / 3 )

= **2 hours**

**b) The count rate of a nuclide fell from 200 counts per second to 12.5 counts per second in 120 minutes.**

**Calculate the half-life period of the nuclide.**

If t ^{1}/_{2 } =x

then

200 –x–>100 –x–>50 –x—>25 –x—>12.5

From 200 to 12.5 =4x =120 minutes

=>x = t ^{1}/_{2} = ( 120 / 4 )

= **30 minutes**

**c) After 6 hours the count rate of a nuclide fell from 240 counts per second to 15 counts per second on the GM tube. Calculate the half-life period of the nuclide.**

If t ^{1}/_{2 } = x

then 240 –x–>120 –x–>60 –x—>30 –x—>15

From 240 to 15 =4x =6 hours

=>x = t ^{1}/_{2} = ( 6 / 4 )= ** 1.5 hours**

**d) Calculate the mass of nitrogen-13 that remain from 2 grams after 6 half-lifes if the half-life period of nitrogen-13 is 10 minutes.**

If t ^{1}/_{2 } = x then:

2 —**x**–>1 –**2x**–>0.5 –**3x**—>0.25 –**4x**–>0.125–**5x**—>0.0625–**6x**—>0.03125

After the 6^{th} half life **0.03125 g** of nitrogen-13 remain.

**e) What fraction of a gas remains after 1hour if its half-life period is 20 minutes?**

If t ^{1}/_{2 } = x then:

then 60 /20 = 3x

1 –x–> ^{1}/_{2} –2x–> ^{1}/_{4} –3x—> ^{1}/_{8}

After the 3^{rd} half-life ** ^{1}/_{8}** of the gas remain

**f) 348 grams of a nuclide A was reduced to 43.5 grams after 270days.Determine the half-life period of the nuclide.**

If t ^{1}/_{2 } = x then:

348 –x–>174 –2x–>87 –3x—>43.5

From 348 to 43.5=3x =270days

=>x = t ^{1}/_{2} = ( 270 / 3 )

= ** 90 days**

**g) How old is an Egyptian Pharaoh in a tomb with 2grams of ^{14}C if the normal ^{14}C in a present tomb is 16grams.The half-life period of ^{14}C is 5600years.**

** **If t ^{1}/_{2 } = x = 5600 years then:

16 –x–>8 –2x–>4 –3x—>2

3x = ( 3 x 5600 )

= 16800years

**h) 100 grams of a radioactive isotope was reduced 12.5 grams after 81days.Determine the half-life period of the isotope.**

** **If t ^{1}/_{2 } = x then:

100 –x–>50 –2x–>25 –3x—>12.5

From 100 to 12.5=3x =81days

=>x = t ^{1}/_{2}

= ( 81 / 3 )

=** 27 days**

A graph of activity against time is called **decay curve.**

A decay curve can be used to determine the half-life period of an isotope since activity decrease at equal time interval to half the original

= **20** minutes

( 50 – 25 ) => ( 40 – 20 ) = ** 20** minutes

Thus t ^{½}_{ }= **20 minutes**

(ii)Why does the graph tend to ‘O’?

**Smaller particle/s will disintegrate /decay to half its original.**

** There can never be ‘O’/zero particles**