Question

A radioactive isotope has a half-life of $T$ years.How long will it take the activity to reduce to $1\%$ of its original value?

• A. $6.64T$
• B. $4.64T$
• C. $6.46T$
• D. $4.46T$

Answer 1

Answer: (A)

$6.64T$

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

$\Rightarrow 2^n=100$

$\Rightarrow n\log 2=\log {10}^2$

$\Rightarrow n=\frac{2\log 10}{\log 2}=\frac{2}{0.301}=6.64$

From $t=nT$ we have $t=6.64T$

Hence, there are $6.64$ half lives of $T$ years spent.

Thus, the time taken is $6.64T$ years.