Question

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

• A. 3.2 Tyr
• B. 4.6 Tyr
• C. 6.6 Tyr
• D. 9.2 Tyr

Answer 1

Answer: (C)

6.6 Tyr

Answer 2

Form the question,
$N=\frac{N_{0}}{100}$ and $t_{1 / 2}=T y r$
So, $\lambda=\frac{0.693}{T}$
Thus, $t=\frac{2.303}{\lambda} \log \left[\frac{N_{0}}{N}\right]$
$=\frac{2.303 T}{0.693} \log 100$
$=6.65\, T$