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Q.

Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is

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a

$\frac{1}{2}$

b

$\frac{1}{3}$

c

$\frac{1}{4}$

d

$\frac{1}{8}$

answer is D.

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Detailed Solution

$n = \frac{72000}{24000} = 3; Now \frac{N}{N_{0}} = {(\frac{1}{2})}^{n} = \frac{1}{8}$

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