**Def **: *Refers to the spontaneous disintegration of certain atomic nuclei which is followed by the emission of alpha particles ( helium nucleus ) , beta particles ( electrons or positrons ) or gamma radiation ( short wavelength electromagnetic waves ).*

Radioactivity is caused by unstable nuclear configurations resulting from nuclear size , limited range of nuclear forces or due to imbalances in small nuclides.

**Measuremen**t –

A_{c} = – dN/dt where N is the number of nuclei present in a radioactive sample at a particular instant. The value is negative as with time the no of elements gets reduced. Unit is Bequeral = 1 disintegration / sec.

Points to note –

*Radioactivity is a random process . The chance of a particular atom in a radioactive sample undergoing disintegration is predicted by applications of probability.**Radioactivity is a nuclear process . Extra-nuclear parts of the atom are not involved in the process.**Radioactivity is not a chemical process because the electronic configuration of the element is not changed. A chemical process involving radioactive elements results in radioactive products.**Radioactive disintegration is uniform throughout the sample.*

**Radioactive Decay Law** –

States that ‘ the activity of a radioactive element at any instant is directly proportional to the number of undecayed active atoms ( parent atoms ) present at that instant’ .

A_{c} = | dN/dt | = λ N , here λ is the decay constant for a decay process .

This equation can be rewritten as dN / dt = – λ N , the negative sign means than dN /dt i.e, the rate at which the active elements are disintegrating decreases with time . Now , integrating with time limits t = 0 to t = t , we arrive at –

**N = N _{0} e ^{-λ t}** – this is the radioactive decay equation , which gives the number of active parent atoms N present at time t in the sample.

λ N = λ N_{0} e ^{– λ t} = ** A _{c} = A_{c0} e ^{– λ t} ** – this equation can be used to find the activity of a radioactive substance at any instant .

**Half – life** – It’s the time required for half the original nuclides to decay , i.e N = N_{0}/2

therefore , N_{0}/2 = N_{0} e ^{– λ t} or In [ 1/2 ] = – λ t or In[ 2 ] = λ t , hence –

** T = In [ 2 ] / λ = 0.693 / λ**

Decay equation in terms of Half – life time –

N = N_{0} (2) ^{– v t }

**Average Life** –

T_{avg} = Sum of the ages of all the nuclei / N_{0} = ** ( ∫ _{0}^{∞} t λ N_{0} e ^{– λ t }) / N_{0}**

Points to Note –

- The number of undecayed nuclei left after n mean life =
**N ( 0.37)**^{n}N_{o}= [ 1/e ]^{n}N_{0 } - Consider a nuclide decaying simultaneously by two different processes having decay constants λ1 and λ2 . In such a case , the effective decay constant of the nuclide = λ = λ
_{1}+ λ_{2},**N = N**_{0}e^{– ( λ1 + λ2 ) t}

On to some problems ** –**

1 . The half life of the radioactive radon is 3.8 days . The time , at the end of which 1/20 th of the radon sample will remain undecayed is ( given log_{10} e = 0.4343 ) ( IIT JEE 1981 )

- (a) 3.8 days
- (b) 16.5 days
- (c) 33 days
- (d) 76 days

Ans . N = N_{0} e ^{-λ t} and λ = In 2 / t _{½} = In (2) / 3.8 . Therefore , N_{0} / 20 = N_{0} e ^{– In (2) / 3.8}

Solving by using the given data gives t = 16.5 days . Hence correct option is (b)

2 . During a negative beta decay – ( IIT JEE 1987 )

- (a) an atomic electron is ejected
- (b) an electron which is already present within the nucleus is ejected
- (c) a neutron in the nucleus decays emitting an electron
- (d) a part of the binding energy of the nucleus is converted into an electron

Ans . The reaction mentioned here is _{0}n^{1 } → _{1}H^{1} + _{-1}e^{0} + antineutrino . Hence correct option is (c)

3 . A freshly prepared radioactive source of half life 2 hr emits radiation of intensity which is 64 times the permissible safety level. The minimum time after which it would be possible to work safely with this source is ( IIT JEE 1988 )

- (a) 6 h
- (b) 12 h
- (c) 24 h
- (d) 128 h

Ans . R = R_{0} [ 1/2 ]^{n} , therefore 1 = 64 [ 1/2 ] ^{n} or n = 6 no of half lives , hence t = n . t_{t½} = 6 x 2 = 12 h. So , correct option is (b)

4 . A small quantity of solution containing Na ^{24} radio nuclide ( half – life = 15 h ) of activity 1.0 microcurie is injected into the blood of a person . A sample of the blood of volume 1 cm³ taken after 5 h shows an activity of 296 disintegrations per minute . Determine the total volume of blood in the body of the person . Assume that the radioactive solution mixes uniformly in the blood of the person. ( 1 curie = 3.7 x 10^{10} disintegrations per second ) ( IIT JEE 1994 )

Ans . Disintegration constant = λ , 0.693 / t_{½}= 0.693 / 15 h = 0.0462 / h

Now , the initial activity is 1 micro curie , i.e , 3.7 x 10^{4} disintegrations / second. Let r denote the activity in 1 cm³ of blood at t = 5h .

= 296 / 60 disintegrations / second = 4.93 disintegrations / second. Let R be the activity of the whole blood of the person at t = 5 h.

So, the total vol. of blood = V = R/r = R_{0} e ^{– λ t} / r = [ 3.7 x 104/4.93 ] e ^{– ( 0.0462 ) ( 5 )} cm³ .

** V = 5.95 L**