First slide

Radio activity

Question

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

Moderate

A

60%
B

65%
C

70%
D

75%

Solution

$N = N_{0} e^{- λt} ∴ 0.9 N_{0} = N_{0} e^{- λ x 5} \Rightarrow 5 λ = \log_{e} \frac{1}{0.9} . . . . . . . . . (i) and {xN}_{0} = N_{0} e^{- λ x 20} \Rightarrow 20 λ = \log_{e} (\frac{1}{x}) . . . . . . . . . (ii)$
Dividing (i) by (ii) we get
$\frac{1}{4} = \frac{\log_{e} (1 / 0.9)}{\log_{e} (1 / x)} = \frac{\log_{10} (1 / 0.9)}{\log_{10} (1 / x)} = \frac{\log_{10} 0.9}{\log_{10} x} \Rightarrow \log_{10} x = 4 \log_{10} 0.9 \Rightarrow x = 0.658 = 65.8 %$

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