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Question

Chemistry Question on Chemical Kinetics

A human body required the 0.01 M activity of radioactive substance after 24 h. Half-life of radioactive substance is 6 h. Then, injection of maximum activity of radioactive substance that can be injected will be

A

0.08

B

0.04

C

0.16

D

0.32

Answer

0.16

Explanation

Solution

Remaining activity = 0.01 M after 24 h
Remaining activity
\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =Initial activity ×(12)n\times\Bigg(\frac{1}{2}\Bigg)^n
Used half-life time(n)=Totaltimet1/2=246=4\frac{Total time}{t_{1/2}}=\frac{24}{6}=4
So \, \, \, \, \, \, \, 0.01 = Initial activity ×(12)4\times\Bigg(\frac{1}{2}\Bigg)^4
Initial activity = 0.01 ×\times 16 = 0.16