The half-life of radium-226 is 1620 years. Th | vedclass

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(D) Given initial amount $N_o = 10 \ g$,final amount $N = 1.25 \ g$,and half-life $t_{1/2} = 1620 \ years$. Using the radioactive decay formula: $N =

Vedclass · Accepted Answer

$4860$

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(D) Given initial amount $N_o = 10 \ g$,final amount $N = 1.25 \ g$,and half-life $t_{1/2} = 1620 \ years$. Using the radioactive decay formula: $N = N_o (1/2)^n$,where $n$ is the number of half-lives $(n = t / t_{1/2})$. Substituting the values: $1.25 = 10 	imes (1/2)^n$. $1.25 / 10 = (1/2)^n \implies 0.125 = (1/2)^n \implies 1/8 = (1/2)^n$. Since $1/8 = (1/2)^3$,we get $n = 3$. Now,calculating the total time: $t = n 	imes t_{1/2} = 3 	imes 1620 = 4860 \ years$.

The half-life of radium-$226$ is $1620$ years. The time taken to convert $10 \ g$ of radium to $1.25 \ g$ is ......... years.

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