Question

Derive an expression for the law of radioactive decay.

Answer 1

Materials which undergo alpha, beta or gamma decays are called radioactive materials. The nature of radioactivity is such that if we have one atom of the radioactive material, we can never predict how long it will take to decay. If we have N₀ number of radioactive atoms (parent atoms or nuclei) of a particular kind, say uranium, at time t = 0, all we can say is that their number will decrease with time as some nuclei (we cannot say which ones) will decay. Let us assume that at time t, number of parent nuclei which are left is N(t). How many of these will decay in the interval between t and t + dt? We can guess that the larger the value of N (t), larger will be the number of decays dN in time dt. Thus, we can say that dN will be proportional to N(t). Also, we can guess that the larger the interval dt, the larger will be the number of particles decaying in that interval. Thus, we can write,

dN ∝ − N(t)dt,

or, dN = − λ N(t)dt  ....(i)

where λ is a constant of proportionality called the decay constant. The negative sign in Eq. (i) indicates that the change in the number of parent nuclei dN, is negative, i.e., N(t) is decreasing with time. We can integrate this equation as

`int_(N_0)^(N(t)) (dN)/(N(t)) = λ int_0^tdt`,

Here, N₀ is the number of parent atoms at time t = 0. Integration gives,

`ln  (N (t))/N_0 = − λt`,

or, N(t) = N₀e^−λt   ....(ii)

This is the decay law of radioactivity. The rate of decay, i.e., the number of decays per unit time `-(dN (t))/dt`, also called the activity A(t), can be written using Eq. (i) and Eq. (ii) as,

A(t) = `- (dN)/(dt) = λ N (t) = λ N_0e^(-λt)`  ....(iii)

At t = 0, the activity is given by A₀ = λN₀.

Using this, Eq. (iii) can be written as,

A(t) = A₀e^−λt   ....(iv)

Activity is measured in units of becquerel (Bq) in SI units. One becquerel is equal to one decay per second. Another unit to measure activity is curie (Ci). One curie is 3.7 × 10¹⁰ decays per second. Thus, 1 Ci= 3.7 × 10¹⁰ Bq.