(a) Derive the mathematical expression for law of radioactive decay for a sample of a radioactive nucleus.
(b) How is the mean life of a given radioactive nucleus related to the decay constant?

Question

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Answer 1

a) According to radioactive decay law, the rate of decay is proportional to amount of nuclei ,N. Thus, $- \frac{d N}{d t} \propto N$ (negative sign come because the amount of nuclei decreases with time)
So, $\frac{d N}{d t} = - λ N$ or $\frac{d N}{N} = - λ t$
Integrating, $l n N = - λ t + c$ where C is the integrating constant.
At $t = 0$ , $N = N_{0}$ so $l n N_{0} = c$
Now, $l n N = - λ t + l n N_{0}$
or $l n (N / N_{0}) = - λ t$ or $N / N_{0} = e^{- λ t}$
Hence, $N = N_{0} e^{- λ t}$ , this is the mathematical expression of radioactive decay of nuclei.
b) The relation between mean life $(t_{a v g})$ and decay constant $(λ)$ : $t_{a v g} = \frac{1}{λ}$

Answer 2

a) According to radioactive decay law, the rate of decay is proportional to amount of nuclei ,N. Thus,

- \frac{d N}{d t} \propto N

(negative sign come because the amount of nuclei decreases with time)
So,

\frac{d N}{d t} = - λ N

or

\frac{d N}{N} = - λ t

Integrating,

l n N = - λ t + c

where C is the integrating constant.
At

t = 0

,

N = N_{0}

so

l n N_{0} = c

Now,

l n N = - λ t + l n N_{0}

or

l n (N / N_{0}) = - λ t

or

N / N_{0} = e^{- λ t}

Hence,

N = N_{0} e^{- λ t}

, this is the mathematical expression of radioactive decay of nuclei.
b) The relation between mean life

(t_{a v g})

and decay constant

(λ)

:

t_{a v g} = \frac{1}{λ}