Solution: Radioactivity of the of the sample, R=λNSo, at t=0:Ao=λNo 1Also, at t=to:Ao3=λN' 2Subtracting equation 1 and 2, we get:Ao-Ao3=λNo-λN'⇒No-N'=2Ao3λ where, λ=ln2T1/2=0 693T1/2

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Solution:
Radioactivity of the of the sample,

R = λ N So, at t = 0 : A_{o} = λ N_{o} . . . . . (1) Also, at t = t_{o} : \frac{A_{o}}{3} = λ N' . . . . . (2) Subtracting equation (1) and (2), we get : A_{o} - \frac{A_{o}}{3} = λ N_{o} - λ N' \Rightarrow N_{o} - N' = \frac{2 A_{o}}{3 λ} (where, λ = \frac{\ln 2}{T_{1 / 2}} = \frac{0.693}{T_{1 / 2}})

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