Consider a radioactive sample which is a mixture of two radioactive species; one with a population $N_{1}$ and mean life of $t_{1}$ and another with a population $N_{2}$ and mean life of $t_{2}$ . Then, the effective mean life of the nuclei in the mixture will be

\frac{t_{1} + t_{2}}{2}

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\frac{t_{1} t_{2}}{t_{1} + t_{2}}

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\frac{(N_{1} t_{1}) (N_{2} t_{2})}{N_{1} + N_{2}}

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\frac{N_{1} t_{1} + N_{2} t_{2}}{N_{1} + N_{2}}

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Solution

The correct option is D $\frac{N_{1} t_{1} + N_{2} t_{2}}{N_{1} + N_{2}}$
$S_{1} \to$ Sum of life of $N_{1}$ nuclei (1st species)
$S_{2} \to$ Sum of life of $N_{2}$ nuclei (2nd species)
$\Rightarrow$
Mean life of mixture = $t = \frac{S_{1} + S_{2}}{N_{1} + N_{2}} = \frac{N_{1} t_{1} + N_{2} t_{2}}{N_{1} + N_{2}}$

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Consider a radioactive sample which is a mixture of two radioactive species; one with a population N1 and mean life of t1 and another with a population N2 and mean life of t2. Then, the effective mean life of the nuclei in the mixture will be

The correct option is D N1t1+N2t2N1+N2S1→ Sum of life of N1 nuclei (1st species) S2→ Sum of life of N2 nuclei (2nd species) ⇒ Mean life of mixture = t=S1+S2N1+N2=N1t1+N2t2N1+N2

Consider a radioactive sample which is a mixture of two radioactive species; one with a population $N_{1}$ and mean life of $t_{1}$ and another with a population $N_{2}$ and mean life of $t_{2}$ . Then, the effective mean life of the nuclei in the mixture will be