Question

A radioactive material reduces to 20% of its initial quantity in 10 hours. find decay constant and half life.
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Answer 1

Answer:

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Explanation:

given:- No of nuclei (N) =100%

dN/dt =20%=remaining no of nuclei

NOW,

(i)

$\frac{dn}{dt} = {lamda} \: n$

${lamda} = 20 \div 100 = 0.2$

(ii)

$half \: life = ln(2) \div lamda = 0.693 \div 0.2 = 3.456 = 3.5 hr = 3hr30min$