1.

A human body required 0.01 M activity of a redioactive substance after 24 hours. Half life of the radioactive substance is 6 hours.Theinjectionof maximum activity of the radioactive substance that can be injected is

Answer»

0.08
0.04
`0.16`
`0.32`

Solution :24 HOURS = 4 half-lives (as `t_(1//2) = 6` hours)
Activity after n half-lives = `([A]_(0))/(2^n)`
`:. 0.01 M = ([A_0])/(2^4) " or " [A_0] = 0.1 xx 16= 0.16 M`.


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