# What is the half-life of the substance if a sample of a radioactive substance decayed to 97.5% of its original amount after a year? (b) How long would it take the sample to decay to 80% of its original amount? _______years??

##### 1 Answer
Mar 29, 2015

(a). ${t}_{\frac{1}{2}} = 27.39 \text{a}$

(b). $t = 8.82 \text{a}$

${N}_{t} = {N}_{0} {e}^{- \lambda t}$

${N}_{t} = 97.5$

${N}_{0} = 100$

$t = 1$

So:

$97.5 = 100 {e}^{- \lambda .1}$

${e}^{- \lambda} = \frac{97.5}{100}$

${e}^{\lambda} = \frac{100}{97.5}$

$\ln {e}^{\lambda} = \ln \left(\frac{100}{97.5}\right)$

$\lambda = \ln \left(\frac{100}{97.5}\right)$

$\lambda = \ln \left(1.0256\right) = 0.0253 \text{/a}$

${t}_{\frac{1}{2}} = \frac{0.693}{\lambda}$

${t}_{\frac{1}{2}} = \frac{0.693}{0.0253} = \textcolor{red}{27.39 \text{a}}$

Part (b):

${N}_{t} = 80$

${N}_{0} = 100$

So:

$80 = 100 {e}^{- 0.0253 t}$

$\frac{80}{100} = {e}^{- 0.0235 t}$

$\frac{100}{80} = {e}^{0.0253 t} = 1.25$

Taking natural logs of both sides:

$\ln \left(1.25\right) = 0.0253 t$

$0.223 = 0.0253 t$

$t = \frac{0.223}{0.0253} = \textcolor{red}{8.82 \text{a}}$