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## Answered question

2022-01-02

Radium decomposes at a rate proportional to the amount present. In 100 years, 100 mg of radium decompose to 96mg. How many mg will be left after another 100 years What is the "half life" (the time required to decompose half the initial amount) of radium?

### Answer & Explanation

Neil Dismukes

Beginner2022-01-03Added 37 answers

We have $t=100yr$, the initial amount of radium $No=100mg$, the final amount of radium id $N=96mg$.
The decay equation for 100 years:
$N={N}_{0}{e}^{-\lambda t}$
$96=100{e}^{-100\lambda }$
${\left(96\right)}^{2}={\left(100{e}^{-100\lambda }\right)}^{2}$
${96}^{2}={10}^{4}{e}^{-200\lambda }$ (A)
Let $\lambda$ be the decay constant.
${N}^{\prime }={N}_{0}{e}^{-200\lambda }$
${N}^{\prime }=100{e}^{-200\lambda }$ (B)
$\frac{{96}^{2}}{{N}^{\prime }}=\frac{{10}^{4}{e}^{-200\lambda }}{10{e}^{-200\lambda }}$
${N}^{\prime }=92.16$
$N={N}_{0}{e}^{-\lambda t}$
$96=100{e}^{-100\lambda }$
$0.96={e}^{-100\lambda }$
$\mathrm{ln}\left(0.96\right)=\mathrm{ln}\left({e}^{-100\lambda }\right)$
$-100\lambda =\mathrm{ln}\left(0.96\right)$
$\lambda =0.0004$
The expression for half-life $\left({T}_{\frac{1}{2}}\right)$ is,
${T}_{\frac{1}{2}}=\frac{0.693}{\lambda }$
$=\frac{0.693}{0.0004}$
$=1732.5$ years

Tiefdruckot

Beginner2022-01-04Added 46 answers

Radium = R.
, k = proportionality constant. =

$\mathrm{ln}\left(R\right)=kt+C$ (1), C is integral constant.
if $t=100$ years, amount of radium = 96mg.
$⇒\mathrm{ln}\left(100\right)=k\left(0\right)+C$
$⇒\mathrm{ln}\left(100\right)=C$
$⇒C=4.6051702$
$\mathrm{ln}\left(R\right)=kt+4.6051702$ (2)
$⇒\mathrm{ln}\left(96\right)=k\left(100\right)+4.6051702$
$⇒\frac{\mathrm{ln}\left(96\right)-4.6051702}{100}=k$
$⇒k=-0.000408219$
$\mathrm{ln}\left(R\right)=-0.000408219t+4.6051702$ (3)
a) Another 100 years
$⇒\mathrm{ln}\left(R\right)=-0.000408219\left(200\right)+4.6051702$
$⇒\mathrm{ln}\left(R\right)=4.523526196$
$⇒R+{e}^{4.523526196}$
$⇒R=92.16mg$
Amount of radium present after another 100 years is 92.16mg.
b): For half-life, $R=50mg$
$⇒\mathrm{ln}\left(50\right)=-0.000408219t+4.6051702$
$⇒t=\frac{\mathrm{ln}\left(50\right)-4.6051702}{-0.000408219}$
$⇒t=1697.97876$ years
$⇒t\approx 1698$ years

karton

Expert2022-01-09Added 613 answers

Decay equation:

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