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Q. The half-life of ${ }_{11}^{24} Na$ is $15.0\, h$. What percentage of it remains after $60 \,h$ ?

Structure of Atom

Solution:

${ }_{11}^{24} Na \rightarrow{ }_{12}^{24} Mg +{ }_{-1}^0 e$
$(\beta)$
$y=$ number of half-life $=\frac{60}{15}=4$
$N=N_0\left(\frac{1}{2}\right)^y$
$=100\left(\frac{1}{2}\right)^4=\frac{100}{16}=6.25 \%$