The activity of a freshly prepared radioactiv | vedclass

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(A) Given: Activity $A = |\frac{dN}{dt}| = 10^{10} \ s^{-1}$.Mean life $	au = 10^9 \ s$.We know that $	au = \frac{1}{\lambda}$,so the decay constant

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(A) Given: Activity $A = |\frac{dN}{dt}| = 10^{10} \ s^{-1}$.Mean life $	au = 10^9 \ s$.We know that $	au = \frac{1}{\lambda}$,so the decay constant $\lambda = \frac{1}{	au} = 10^{-9} \ s^{-1}$.From the law of radioactive decay,$A = \lambda N$,where $N$ is the number of atoms.$10^{10} = 10^{-9} 	imes N \implies N = 10^{19}$ atoms.The mass of one atom is $m_a = 10^{-25} \ kg$.Total mass $M = N 	imes m_a = 10^{19} 	imes 10^{-25} \ kg = 10^{-6} \ kg$.Converting to milligrams $(mg)$: $10^{-6} \ kg = 10^{-3} \ g = 1 \ mg$.

The activity of a freshly prepared radioactive sample is $10^{10}$ disintegrations per second,whose mean life is $10^9 \ s$. The mass of an atom of this radioisotope is $10^{-25} \ kg$. The mass (in $mg$) of the radioactive sample is

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