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Question

To determine the half life of a radioactive element a student plots a graph of lndN(t)dt versus t. Here dN(t)dt is the rate of radio active decay at time t. If the number of radio active nuclei of this element decreases by a factor of p after 4.16years, the value p is:
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Solution

N=Noeλt[Radioactive nucliei of the element which is decaying after time t]
dNdt=λNoeλt
lndNdt=lnλNoλt
By graph
λ=12
Nfinal=NoP=Noeλt
P=eλt
=e12×4.16
=e2.08
P8

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