Question

A radioactive isotope decomposes according to the first order with half-life period of 15 hrs. 80% of the sample will decompose in:

• A. $1.5\times 0.8hr$
• B. $15\times \left(log8\right)hr.$
• C. $34.83hr.$
• D. none of these

Answer 1

Answer: (C)

$34.83hr.$

$k=\frac{0.693}{15}$ = $0.0462{hr}^{-1}$

$t=\frac{2.030}{k}log\frac{a}{a-x}$

Here 80% of the sample will decompose thus,

$t=\frac{2.303}{k}log\frac{100}{100-80}$

$t=\frac{2.303}{0.0462}log5$

$t=34.83hr$

So, Option C is correct