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??

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Answer 1 · 2015-03-29T22:35:29

(a). $t_{\frac{1}{2}} = 27.39 a$ $t_(1/2)=27.39"a"$

(b). $t = 8.82 a$ $t=8.82"a"$

$N_{t} = N_{0} e^{- λ t}$ $N_t=N_0e^(-lambda t)$

$N_{t} = 97.5$ $N_t=97.5$

$N_{0} = 100$ $N_0=100$

$t = 1$ $t=1$

So:

$97.5 = 100 e^{- λ .1}$ $97.5=100e^(-lambda.1)$

$e^{- λ} = \frac{97.5}{100}$ $e^(-lambda)=(97.5)/(100)$

$e^{λ} = \frac{100}{97.5}$ $e^(lambda)=(100)/(97.5)$

${ln e}^{λ} = ln (\frac{100}{97.5})$ $lne^(lambda)=ln((100)/(97.5))$

$λ = ln (\frac{100}{97.5})$ $lambda=ln((100)/(97.5))$

$λ = ln (1.0256) = 0.0253 /a$ $lambda=ln(1.0256)=0.0253"/a"$

$t_{\frac{1}{2}} = \frac{0.693}{λ}$ $t_((1)/(2))=0.693/lambda$

$t_{\frac{1}{2}} = \frac{0.693}{0.0253} = 27.39 a$ $t_((1)/(2))=0.693/0.0253=color(red)(27.39"a")$

Part (b):

$N_{t} = 80$ $N_t=80$

$N_{0} = 100$ $N_0=100$

So:

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

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

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

Taking natural logs of both sides:

$ln (1.25) = 0.0253 t$ $ln(1.25)=0.0253t$

$0.223 = 0.0253 t$ $0.223=0.0253t$

$t = \frac{0.223}{0.0253} = 8.82 a$ $t=0.223/0.0253=color(red)(8.82"a")$