The half-life of radium is 1620 years and its | vedclass

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(A) The rate of decay is given by $\frac{dN}{dt} = \lambda N$. First,calculate the decay constant $\lambda$ in $s^{-1}$: $\lambda = \frac{0.693}{T_{1/

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$3.61 	imes 10^{10}$

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(A) The rate of decay is given by $\frac{dN}{dt} = \lambda N$. First,calculate the decay constant $\lambda$ in $s^{-1}$: $\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1620 	imes 365 	imes 24 	imes 3600} \ s^{-1}$. Next,calculate the number of atoms $N$ in $1 \ g$ $(10^{-3} \ kg)$ of radium: $N = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A = \frac{10^{-3} \ kg}{226 \ kg/kmol} 	imes 6.02 	imes 10^{26} \ atoms/kmol = \frac{6.02}{226} 	imes 10^{23} \ atoms$. Now,calculate the activity $\frac{dN}{dt}$: $\frac{dN}{dt} = \left( \frac{0.693}{1620 	imes 3.1536 	imes 10^7} ight) 	imes \left( \frac{6.02}{226} 	imes 10^{23} ight) \approx 3.61 	imes 10^{10} \ atoms/s$.

The half-life of radium is $1620$ years and its atomic weight is $226 \ kg/kmol$. The number of atoms that will decay from its $1 \ g$ sample per second will be (Avogadro's number $N_A = 6.02 \times 10^{26} \ atoms/kmol$)

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