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

Question

A human body required the 0.01 M activity of radioactive substance after 24 h. Half-life of radioactive substanu is 6 h. Then injection of maximum activity of radioactive substance that can be injected (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 ×((1/2))n Used half- life time ( n ) =( text Total time /T1 / 2)=(24/6)=4 So, 0.01= Initial activity ×((1/2))4 Initial activity =0.01 × 16=0.16

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

0.08

0.04

0.16

0.32

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

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