A radio active substance has a life cycle of four months. Three fourth of the substance will decay in:

three months

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four months

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eight months

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twelve months

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Solution

The correct option is B eight months
It is given that a radioactive has a life cycle of four months.
Substance left undecayed, $N_{0} - \frac{3}{4} N_{0} = \frac{1}{4} N_{0}$
$\frac{N}{N_{0}} = \frac{1}{4} = {(\frac{1}{2})}^{n}$
Here, $n = 2$
$∴$ Three fourth of the substance will decay in $t = n T = 2 \times 4 = 8$ months

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A radio active substance has a life cycle of four months. Three fourth of the substance will decay in:

The correct option is B eight monthsIt is given that a radioactive has a life cycle of four months.Substance left undecayed, N0−34N0=14N0NN0=14=(12)nHere, n=2∴ Three fourth of the substance will decay in t=nT=2×4=8 months