What is the mass of 1 , text{curie} of U^{234 | vedclass

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Question

(C) The activity of $1 \, 	ext{curie}$ is defined as $A = 3.7 	imes 10^{10} \, 	ext{disintegrations/second}$.The decay constant $\lambda$ for $U^{2

Vedclass · Accepted Answer

$1.48 	imes 10^{-11} \, 	ext{g}$

Vedclass · Answer

(C) The activity of $1 \, 	ext{curie}$ is defined as $A = 3.7 	imes 10^{10} \, 	ext{disintegrations/second}$.The decay constant $\lambda$ for $U^{234}$ is given by $\lambda = \frac{\ln 2}{T_{1/2}}$. The half-life $T_{1/2}$ of $U^{234}$ is approximately $2.455 	imes 10^5 \, 	ext{years} = 2.455 	imes 10^5 	imes 3.1536 	imes 10^7 \, 	ext{s} \approx 7.74 	imes 10^{12} \, 	ext{s}$.Using the law of radioactivity, $A = \lambda N$, where $N$ is the number of atoms.$N = \frac{A}{\lambda} = \frac{A 	imes T_{1/2}}{\ln 2} = \frac{3.7 	imes 10^{10} 	imes 7.74 	imes 10^{12}}{0.693} \approx 4.13 	imes 10^{23} \, 	ext{atoms}$.The mass $m$ is given by $m = \frac{N 	imes M}{N_A}$, where $M = 234 \, 	ext{g/mol}$ and $N_A = 6.022 	imes 10^{23} \, 	ext{atoms/mol}$.$m = \frac{4.13 	imes 10^{23} 	imes 234}{6.022 	imes 10^{23}} \approx 160.5 \, 	ext{g}$.Note: The provided option $C$ $(1.48 	imes 10^{-11} \, 	ext{g})$ is a common textbook error resulting from treating the activity value as the number of atoms directly. Following standard physics calculation, the mass is significantly higher.

What is the mass of $1 \, \text{curie}$ of $U^{234}$?

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