The decay constant of Ra^{226} is 1.37 times | vedclass

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Question

(A) The activity $A$ is given by the formula $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of atoms.Given: $\lambda = 1.

Vedclass · Accepted Answer

$4.1 	imes 10^{18}$

Vedclass · Answer

(A) The activity $A$ is given by the formula $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of atoms.Given: $\lambda = 1.37 	imes 10^{-11} \ s^{-1}$ and $A = 1.5 \ mCi$.Since $1 \ mCi = 3.7 	imes 10^7 \ 	ext{disintegrations per second (dps)}$,$A = 1.5 	imes 3.7 	imes 10^7 \ dps = 5.55 	imes 10^7 \ dps$.Using $N = \frac{A}{\lambda}$,we get:$N = \frac{5.55 	imes 10^7}{1.37 	imes 10^{-11}} \approx 4.05 	imes 10^{18}$ atoms.Rounding to the nearest option,the value is $4.1 	imes 10^{18}$.

The decay constant of $Ra^{226}$ is $1.37 \times 10^{-11} \ s^{-1}$. $A$ sample of $Ra^{226}$ having an activity of $1.5 \ mCi$ will contain ...... atoms.

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