Question

Because of its chemical similarity to calcium, 90/38 Sr can collect in the bones and present a

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asked 2021-05-08

Because of its chemical similarity to calcium, \(\frac{90}{38}\) Sr can collect in the bones and present a health risk.
Part A
What percentage of \(\frac{90}{38}\) Sr present initially still exists after a period of 46.0y ?
Part B
What percentage of \(\frac{90}{38}\) Sr present initially still exists after a period of 59.0y ?
Part C
What percentage of \(\frac{90}{38}\) Sr present initially still exists after a period of 70.0y ?

Answers (1)

2021-05-10

time \(t = 46\) years
Exponential behavour of the number of undecayed nuclei
\(\displaystyle{N}={N}_{{o}}{e}^{{-\lambda}}{t}\)
\(\displaystyle{\frac{{{N}}}{{{N}_{{o}}}}}={e}^{{-\lambda}}{t}\)
half life of Sr is 28.9 years
decay constant \(\displaystyle\lambda={\frac{{{0.693}}}{{{t}^{{\frac{{1}}{{2}}}}}}}\)
\(=\frac{0.693}{28.9}\)
\(=0.0239\)
\(\displaystyle{\frac{{{N}}}{{{N}_{{o}}}}}={e}^{{-{0.0239}\cdot{46}}}\)
\(\displaystyle={0.3331}\)
\(=33.31\%\)
time \(t = 59\) years
\(\displaystyle{\frac{{{N}}}{{{N}_{{o}}}}}={e}^{{-{0.0239}\cdot{59}}}\)
\(\displaystyle={0.2441}\)
\(=24.41\%\)
time \(t=70\) years
\(\displaystyle{\frac{{{N}}}{{{N}_{{o}}}}}={e}^{{-{0.0239}\cdot{70}}}\)
\(=0.1877\)
\(=18.77\%\)

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