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Question

The initial activity of a certain radioactive isotopes was measured as $$16000$$ counts $$min^{-1}$$. Given that the only activity measured was due to this isotopes and that its activity after $$12 h$$ was $$2100$$ counts $$min^{-1}$$, its half-life, in hours, is nearest to [Given $$log_e(7.2)=2]$$


A
9.0
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B
6.0
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C
4.0
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D
3.0
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Solution

The correct option is C $$4.0$$
Activity $$=\cfrac{dN}{dt} = N_o \lambda e^{- \lambda t}$$
Let initial activity be, $$A_o=16000 min^{-1} $$
$$A(t) =$$ $$A_0 e^{- \lambda t} $$
At $$t = 12\ hr$$
$$ 2100 = 16000 e^{-12 \lambda } $$
$$ \lambda = 0.169 hr^{-1} $$
Half life: $$ t_{1/2} = \cfrac{ln 2}{\lambda} = 4\ hrs $$

Physics
NCERT
Standard XII

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