Question

How long will a radioactive isotope whose half life is t years take for its activity to reduce to 1/8 of its initial value?

Answer 1

2t= 1
Which means in 2t years the isotope will completely degraded.
for 1 / 8 of its initial value is 2t / 8
which is t / 4 years.

Answer 2

It will take 3t years for its activity to reduce to 1/8 of its initial value.

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}$

$t=\frac{0.69}{k}$

$k=\frac{0.69}{t}$

Expression for rate law for first order kinetics is given by:

$T=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant = 1

a - x = amount left after decay process =$\frac{1}{8}$

$T=\frac{2.303}{k}\log\frac{1}{\frac{1}{8}}$

$T=\frac{2.303}{\frac{0.69}{t}}\log\frac{1}{\frac{1}{8}}}$

$T=3t$

Thus it will take 3t years for its activity to reduce to 1/8 of its initial value.

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