Question

Select One Option Correct from the following : The decay constant of a radioactive sample is $\lambda$ The half-life and mean-life of the sample are respectively given by
(A)$1/\lambda and(\ln 2)/\lambda$
(B)$(\ln 2)/\lambda and 1/\lambda$
(C) $\lambda(\ln 2) and 1/ \lambda$
(D)$\lambda / (\ln 2) and 1/ \lambda$

Answer 1

When a radioactive material undergoes α, β or γ-decay, the number of nuclei undergoing the decay, per unit time, is proportional to the total number of nuclei in the sample material. So,

If N = total number of nuclei in the sample and ΔN = number of nuclei that undergo decay in time Δt then,

ΔN/ Δt ∝ N

Or, ΔN/ Δt = λN … (1)

where λ = radioactive decay constant or disintegration constant. Now, the change in the number of nuclei in the sample is, dN = – ΔN in time Δt. Hence, the rate of change of N (in the limit Δt→ 0) is,

dN/dt = – λN

Or, dN/N = – λdt

Now, integrating both sides of the above equation, we get,

NN0∫ dN/N = λ tt0∫ dt … (2)

Or, ln N – ln N0 = – λ (t – t0) … (3)

Where, N0 is the number of radioactive nuclei in the sample at some arbitrary time t0 and N is the number of radioactive nuclei at any subsequent time t. Next, we set t0 = 0 and rearrange the above equation (3) to get,

ln (N/N0) = – λt

Or, N(t) = N0e– λt … (4)

Equation (4) is the Law of Radioactive Decay