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Question
Obtain an expression for the decay law of radioactivity. Hence show that the activity A(t) =λNO e-λt.
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Solution
Expression for a decay law of radioactivity:
If ‘N(t)’ is the number of parent nuclei present at any instant ‘t’, ‘dN’ is the number of nuclei disintegrated in a short interval of time ‘dt’, then,
dN ∝ −N(t) dt
dN = −λ N(t) dt
where λ is known as decay constant or disintegration constant. The negative sign indicates the disintegration of atoms.
Integrating both sides of the equation,
`int_{"N"_0}^{"N"_(("t"))} "dN"/("N"("t")) = int_0^"t" -lambda "dt"`
where, N0 is number of parent atoms at time t = 0.
∴ loge = `("N"("t"))/("N"_0) = -lambda"t"`
∴ N(t) = `"N"_0 "e"-lambda"t"`
Expression for activity:
- The rate of decay, i.e., the number of decays per unit time −`("dN"("t"))/"dt"`, is called as activity A(t).
- It is given as,
A(t) = −`("dN"("t"))/"dt"`
= λN(t)
= λN0e−λt
At t = 0, the activity is A0 = λN0.
∴ A(t) = A0e−λt
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