The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $(1 / 16)^{th}$ of its initial value?

Option 1 - 80 minutes

Option 2 - 20 minutes

Option 3 - 40 minutes

Option 4 - 60 minutes

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Answered by

Alok Kumar Singh | Contributor-Level 10

5 months ago

Correct Option - 2

Detailed Solution:

$\frac{1}{2} m {(\frac{u}{3})}^{2} - \frac{1}{2} {m u}^{2} = - F_{R} * 24$

$0 - \frac{1}{2} {m u}^{2} = - F_{R} * d$

$\frac{\frac{1}{2} {m u}^{2}}{\frac{1}{2} m u^{2} * \frac{8}{9}} = \frac{d}{24}$

$d = 24 * \frac{9}{8} = 27 c m$

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The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to ( 1 / 16 ) th of its initial value?

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The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $(1 / 16)^{th}$ of its initial value?