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If 10% of a radioactive material decays in 5 days then the amount of the original material left after 20 days is approximately

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60 %

65 %

70 %

75 %

answer is B.

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Detailed Solution

$N = \frac{N_{0}}{2^{t / T_{1 / 2}}}$

$90 = \frac{100}{2^{5 / T_{1 / 2}}} \Rightarrow 2^{5 / T_{1 / 2}} = \frac{100}{90}$

$N = \frac{100}{{(2^{20 / T_{1 / 2}})}^{4}} = \frac{100}{{(\frac{100}{90})}^{4}} = \frac{100 \times 81 \times 81}{100 \times 100}$

$N = 65 %$

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