The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8} t h$ of its original mass to disintegrate will be:

Option 1 -

120 days

Option 2 -

130 days

Option 3 -

180 days

Option 4 -

20 days

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

$t_{\frac{1}{2}} = 60 d a y s$

at t = 0 $\frac{m}{m_{0}} \to$ 0

$a t t \to \frac{m}{m_{0}} (1 - \frac{7}{8})$

= $\frac{m}{8 m_{0}} \to \frac{7}{8} (\frac{m}{m_{0}})$

$\Rightarrow \frac{1}{2^{3}} = \frac{1}{2^{(\frac{t}{60})}} \Rightarrow \frac{t}{60} = 3$

t = 180 days

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The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8} t h$ of its original mass to disintegrate will be:

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The half life period of a radioactive substance is 60 days. The time taken for 7 8 t h of its original mass to disintegrate will be:

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The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8} t h$ of its original mass to disintegrate will be: