The half-life of _{38}^{90} Sr is 28 years. W | vedclass

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Question

(A) Half-life of $_{38}^{90} Sr$, $t_{1/2} = 28 \; 	ext{years}$.Converting to seconds: $t_{1/2} = 28 	imes 365 	imes 24 	imes 3600 \approx 8.83

Vedclass · Accepted Answer

$7.878 	imes 10^{10} \; 	ext{atoms/s}$

Vedclass · Answer

(A) Half-life of $_{38}^{90} Sr$, $t_{1/2} = 28 \; 	ext{years}$.Converting to seconds: $t_{1/2} = 28 	imes 365 	imes 24 	imes 3600 \approx 8.83 	imes 10^{8} \; s$.Mass of the isotope, $m = 15 \; mg = 15 	imes 10^{-3} \; g$.The number of atoms $N$ in $15 \; mg$ is given by $N = \frac{m}{M} 	imes N_A$, where $M = 90 \; g/mol$ and $N_A = 6.023 	imes 10^{23} \; 	ext{atoms/mol}$.$N = \frac{15 	imes 10^{-3}}{90} 	imes 6.023 	imes 10^{23} \approx 1.0038 	imes 10^{20} \; 	ext{atoms}$.The decay constant $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{8.83 	imes 10^{8}} \; s^{-1}$.The rate of disintegration is $R = \lambda N = \left( \frac{0.693}{8.83 	imes 10^{8}} ight) 	imes (1.0038 	imes 10^{20}) \approx 7.878 	imes 10^{10} \; 	ext{atoms/s}$.

The half-life of $_{38}^{90} Sr$ is $28$ years. What is the disintegration rate of $15 \; mg$ of this isotope?

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