The half-life of an isotope is 10 hrs. How m | vedclass

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(B) The initial amount $N_0 = 1 \ g$. Given the molar mass $M = 1 \ g/mol$,the initial number of atoms is $N_0 = (1 \ g / 1 \ g/mol) 	imes 6.022 	im

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$4.56 	imes 10^{23} \ 	ext{atoms}$

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(B) The initial amount $N_0 = 1 \ g$. Given the molar mass $M = 1 \ g/mol$,the initial number of atoms is $N_0 = (1 \ g / 1 \ g/mol) 	imes 6.022 	imes 10^{23} \ 	ext{atoms/mol} = 6.022 	imes 10^{23} \ 	ext{atoms}$.Using the radioactive decay formula $N = N_0 	imes (1/2)^{(t/T_{1/2})}$,where $t = 4 \ hrs$ and $T_{1/2} = 10 \ hrs$.$N = 6.022 	imes 10^{23} 	imes (0.5)^{(4/10)} = 6.022 	imes 10^{23} 	imes (0.5)^{0.4}$.Since $(0.5)^{0.4} \approx 0.7578$,we get $N \approx 6.022 	imes 10^{23} 	imes 0.7578 \approx 4.56 	imes 10^{23} \ 	ext{atoms}$.

The half-life of an isotope is $10 \ hrs$. How much will be left behind after $4 \ hrs$ in a $1 \ gm$ sample? (Assume the molar mass of the isotope is $1 \ g/mol$ for calculation purposes).

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