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Q. The half-life period of a radioactive substance is 45 minutes. The time during which 99.9 % of it will disintegrate is

Answer 1

Given:

The half-life period of a radioactive substance is 45 minutes.

To find:

The time during which 99.9 % of it will disintegrate is ?

Calculation:

$\therefore$ Half life = 45 mins.

$\implies \: t_{ \frac{1}{2} } = 45$

$\implies \: \dfrac{0.69}{k} = 45 \: \: \: \: \: .......(1)$

Now , for 99.9% disintegration:

$\therefore \: t = \dfrac{1}{k} ln( \dfrac{ A_{0}}{A} )$

$\implies \: t = \dfrac{1}{k} ln \bigg \{\dfrac{ A_{0}}{A_{0}(1 - 99.9\%)} \bigg \}$

$\implies \: t = \dfrac{1}{k} ln \bigg \{\dfrac{ 1}{(1 - 0.999)} \bigg \}$

$\implies \: t = \dfrac{1}{k} ln \bigg \{\dfrac{ 1}{(0.001)} \bigg \}$

$\implies \: t = \dfrac{1}{k} ln \bigg \{\dfrac{ 1}{ {10}^{ - 3} } \bigg \}$

$\implies \: t = \dfrac{1}{k} ln \bigg \{ {10}^{3} \bigg \}$

$\implies \: t = \dfrac{2.303}{k} \log \bigg \{ {10}^{3} \bigg \}$

$\implies \: t = \dfrac{2.303}{k} \times 3$

$\implies \: t = \dfrac{6.9}{k}$

$\implies \: t = \dfrac{0.69}{k} \times 10$

$\implies \: t = 45 \times 10$

$\implies \: t = 450 \: min$

So, required time is 450 minutes.