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

Question

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

Tardigrade Admin · Accepted Answer

Remaining activity = 0.01 M after 24 h
Remaining activity

=Initial activity

\times (\frac{1}{2})^{n}

Used half-life time(n)=

\frac{T o t a lt im e}{t _{1/2}} = \frac{24}{6} = 4

So

0.01 = Initial activity

\times (\frac{1}{2})^{4}

Initial activity = 0.01

\times

16 = 0.16

Q. 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

0.08

0.04

0.16

0.32

Remaining activity = 0.01 M after 24 h Remaining activity =Initial activity ×(21​)n Used half-life time(n)=t1/2​Totaltime​=624​=4 So 0.01 = Initial activity ×(21​)4 Initial activity = 0.01 × 16 = 0.16

Remaining activity = 0.01 M after 24 h
Remaining activity
=Initial activity $\times (\frac{1}{2})^{n}$
Used half-life time(n)= $\frac{T o t a lt im e}{t _{1/2}} = \frac{24}{6} = 4$
So 0.01 = Initial activity $\times (\frac{1}{2})^{4}$
Initial activity = 0.01 $\times$ 16 = 0.16