A radioactive nuclide X decays at the rate of | vedclass

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(C) The activity of $1 \ g$ of Radium is defined as $1 \ Curie (Ci) = 3.70 	imes 10^{10} \ disintegration \ s^{-1}$.Given,the activity of nuclide $X$

Vedclass · Accepted Answer

$0.00270$

Vedclass · Answer

(C) The activity of $1 \ g$ of Radium is defined as $1 \ Curie (Ci) = 3.70 	imes 10^{10} \ disintegration \ s^{-1}$.Given,the activity of nuclide $X$ is $A_X = 1.00 	imes 10^5 \ disintegration \ s^{-1} \ g^{-1}$.To find the activity in $Ci \ g^{-1}$,we divide the activity of $X$ by the activity of $1 \ g$ of Radium:$Activity (in \ Ci \ g^{-1}) = \frac{1.00 	imes 10^5}{3.70 	imes 10^{10}} = 0.270 	imes 10^{-5} \ Ci \ g^{-1}$.Since $1 \ Ci = 1000 \ mCi$,the activity in $mCi \ g^{-1}$ is $0.270 	imes 10^{-5} 	imes 10^3 = 0.00270 \ mCi \ g^{-1}$.Therefore,the correct option is $C$.

$A$ radioactive nuclide $X$ decays at the rate of $1.00 \times 10^5 \ disintegration \ s^{-1} \ g^{-1}$. Radium decays at the rate of $3.70 \times 10^{10} \ disintegration \ s^{-1} \ g^{-1}$. The activity of $X$ in millicuries $g^{-1}$ $(mCi \ g^{-1})$ is

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