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.
Ac = – 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’ .
Ac = | 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 = N0 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 N0 e – λ t = Ac = Ac0 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 = N0/2
therefore , N0/2 = N0 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 = N0 (2) – v t
Average Life –
Tavg = Sum of the ages of all the nuclei / N0 = ( ∫0∞ t λ N0 e – λ t ) / N0
Points to Note –
- The number of undecayed nuclei left after n mean life = N ( 0.37)n No = [ 1/e ]n N0
- 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 = λ1 + λ2 , N = N0 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 log10 e = 0.4343 ) ( IIT JEE 1981 )
- (a) 3.8 days
- (b) 16.5 days
- (c) 33 days
- (d) 76 days
Ans . N = N0 e -λ t and λ = In 2 / t ½ = In (2) / 3.8 . Therefore , N0 / 20 = N0 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 0n1 → 1H1 + -1e0 + 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 = R0 [ 1/2 ]n , therefore 1 = 64 [ 1/2 ] n or n = 6 no of half lives , hence t = n . tt½ = 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 1010 disintegrations per second ) ( IIT JEE 1994 )
Ans . Disintegration constant = λ , 0.693 / t½= 0.693 / 15 h = 0.0462 / h
Now , the initial activity is 1 micro curie , i.e , 3.7 x 104 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 = R0 e – λ t / r = [ 3.7 x 104/4.93 ] e – ( 0.0462 ) ( 5 ) cm³ .
V = 5.95 L