First slide

Radio activity

Question

Half-life of a radioactive substance is 20 minutes. Difference between points of time when it is 33%
disintegrated and 67% disintegrated is approximate

Moderate

A

10 min
B

20 min
C

30 min
D

40 min

Solution

$λ = \frac{0.693}{T_{1 / 2}} = \frac{0.693}{20} = 0.03465$
Now time of decay $t = \frac{2.303}{λ} \log \frac{N_{0}}{N}$
$\Rightarrow t_{1} = \frac{2.303}{0.03465} \log \frac{100}{67} = 11.6 \min and t_{2} = \frac{2.303}{0.03465} \log \frac{100}{33} = 32 \min$
Thus time difference between points of time
$= t_{1} - t_{2} = 32 - 11.6 = 20.4 \min ≃ 20 \min .$

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