Radium decomposes at a rate proportional to the amount present.

ajedrezlaproa6j 2022-01-02 Answered
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?

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Neil Dismukes
Answered 2022-01-03 Author has 3487 answers
From given, the time is \(\displaystyle{t}={100}{y}{r}\), the initial amount of radium is \(\displaystyle{N}{o}={100}{m}{g}\) and the final amount of radium id \(\displaystyle{N}={96}{m}{g}\).
The decay equation can be written for 100 years.
\(\displaystyle{N}={N}_{{{0}}}{e}^{{-\lambda{t}}}\)
\(\displaystyle{96}={100}{e}^{{-{100}\lambda}}\)
\(\displaystyle{\left({96}\right)}^{{{2}}}={\left({100}{e}^{{-{100}\lambda}}\right)}^{{{2}}}\)
\(\displaystyle{96}^{{{2}}}={10}^{{{4}}}{e}^{{-{200}\lambda}}\) (A)
Let \(\displaystyle\lambda\) be the decay constant.
The decay equation can be written for 100 years.
\(\displaystyle{N}'={N}_{{{0}}}{e}^{{-{200}\lambda}}\)
\(\displaystyle{N}'={100}{e}^{{-{200}\lambda}}\) (B)
Divide A by B.
\(\displaystyle{\frac{{{96}^{{{2}}}}}{{{N}'}}}={\frac{{{10}^{{{4}}}{e}^{{-{200}\lambda}}}}{{{10}{e}^{{-{200}\lambda}}}}}\)
\(\displaystyle{N}'={92.16}\)
\(\displaystyle{N}={N}_{{{0}}}{e}^{{-\lambda{t}}}\)
\(\displaystyle{96}={100}{e}^{{-{100}\lambda}}\)
\(\displaystyle{0.96}={e}^{{-{100}\lambda}}\)
\(\displaystyle{\ln{{\left({0.96}\right)}}}={\ln{{\left({e}^{{-{100}\lambda}}\right)}}}\)
\(\displaystyle-{100}\lambda={\ln{{\left({0.96}\right)}}}\)
\(\displaystyle\lambda={0.0004}\)
The expression for half-life \(\displaystyle{\left({T}_{{{\frac{{{1}}}{{{2}}}}}}\right)}\) is,
\(\displaystyle{T}_{{{\frac{{{1}}}{{{2}}}}}}={\frac{{{0.693}}}{{\lambda}}}\)
\(\displaystyle={\frac{{{0.693}}}{{{0.0004}}}}\)
\(\displaystyle={1732.5}\) years
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Tiefdruckot
Answered 2022-01-04 Author has 2049 answers
Let us assume the amount of radium present be R.
\(\displaystyle{\frac{{{d}{R}}}{{{\left.{d}{t}\right.}}}}={k}{R}\) where k is proportionality constant.
Integrate with respect to t,
\(\displaystyle\int{\frac{{{d}{R}}}{{{R}}}}=\int{k}{\left.{d}{t}\right.}\)
\(\displaystyle{\ln{{\left({R}\right)}}}={k}{t}+{C}\) (1) where C is integral constant.
According to question, at present \(\displaystyle{\left({t}={0}\right)}\) amount of radium present 100mg.
When \(\displaystyle{t}={100}\) years, amount of radium presented 96mg.
Plugging \(\displaystyle{t}={0},{R}={100}\) in equation (1),
\(\displaystyle\Rightarrow{\ln{{\left({100}\right)}}}={k}{\left({0}\right)}+{C}\)
\(\displaystyle\Rightarrow{\ln{{\left({100}\right)}}}={C}\)
\(\displaystyle\Rightarrow{C}={4.6051702}\)
Equation (1) becomes,
\(\displaystyle{\ln{{\left({R}\right)}}}={k}{t}+{4.6051702}\) (2)
Plugging \(\displaystyle{t}={100},{R}={96}\) in equation (2),
\(\displaystyle\Rightarrow{\ln{{\left({96}\right)}}}={k}{\left({100}\right)}+{4.6051702}\)
\(\displaystyle\Rightarrow{\frac{{{\ln{{\left({96}\right)}}}-{4.6051702}}}{{{100}}}}={k}\)
\(\displaystyle\Rightarrow{k}=-{0.000408219}\)
Equation (2) becomes,
\(\displaystyle{\ln{{\left({R}\right)}}}=-{0.000408219}{t}+{4.6051702}\) (3)
Part (a): Another 100 years (i.e., \(\displaystyle{t}={200}\) years)
\(\displaystyle\Rightarrow{\ln{{\left({R}\right)}}}=-{0.000408219}{\left({200}\right)}+{4.6051702}\)
\(\displaystyle\Rightarrow{\ln{{\left({R}\right)}}}={4.523526196}\)
\(\displaystyle\Rightarrow{R}+{e}^{{{4.523526196}}}\)
\(\displaystyle\Rightarrow{R}={92.16}{m}{g}\)
Amount of radium present after another 100 years is 92.16mg.
Part (b): For half-life, \(\displaystyle{R}={50}{m}{g}\)
Plugging in equation (3),
\(\displaystyle\Rightarrow{\ln{{\left({50}\right)}}}=-{0.000408219}{t}+{4.6051702}\)
\(\displaystyle\Rightarrow{t}={\frac{{{\ln{{\left({50}\right)}}}-{4.6051702}}}{{-{0.000408219}}}}\)
\(\displaystyle\Rightarrow{t}={1697.97876}\) years
\(\displaystyle\Rightarrow{t}\approx{1698}\) years
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karton
Answered 2022-01-09 Author has 8454 answers

Decay equation:
\(\begin{array}{}N=N_{0}e^{- \lambda t} \\N_{0}=100 \\N=96 \\t=100\ years \\96=100e^{-\lambda \times 100} \\0.96=e^{-100 \lambda} \\\text{taking log} \\\ln (0.96)=-100 \lambda \\\lambda=-\frac{\ln (0.96)}{100} \\e^{-\lambda}=e^{\frac{\ln (0.96)}{100}} \\=0.9996 \\N=\frac{N_{0}}{2}=50 \\50=100(e^{-\lambda})^{1/2} \\=100 \times (0.9996)^{1/2} \\\frac{1}{2}=(0.9996) \\\text{taking log} \\\ln (\frac{1}{2})=\ln (0.9996) \\t=\frac{\ln (1/2)}{\ln (0.9996)} \\=1732.52 \end{array}\)

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