A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to (a) 3.125%, (b) 1% of its original value?

Solution

Given, the sample has a half life of T years.

(a) The fraction of the original sample left is ,
$\frac{N}{N_{o}}$ = $\frac{3.125}{100}$
$= \frac{1}{32} = {(\frac{1}{2})}^{5}$

Hence, there are 5 half lives of T years spent. Thus, the time taken is 5T years.

(b) The fraction of the original sample left = $\frac{1}{100} = {(\frac{1}{2})}^{n}$

i.e., $2^{n}$ = 100

$\Rightarrow$ n log 2 = log 100

Hence, n = $\frac{\log 100}{\log 2} = \frac{2}{0.301}$ = 6.64

From, t= nT we have, t = 6.64 T.

Hence, there are 6.64 half lives of T years spent. Thus, the time taken is 6.64T years.

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Wave Optics

A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to (a) 3.125%, (b) 1% of its original value?

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