Question

$1\, mg$ radium has $2.68 \times 10^{18}$ atoms. Its half life is $1620$ years. How many radium atoms will disintegrate from $1 \,mg$ of pure radium in $3240$ years?

• A. $2.01\times10^{9}$
• B. $2.01\times10^{18}$
• C. $1.01\times10^{9}$
• D. $1.01\times10^{18}$

Answer 1

Answer: (B)

$n=\frac{t}{T_{1 /2}}=\frac{3240}{1620}=2$
As $\frac{N}{N_{0}}=\frac{m}{m_{0}}=\left(\frac{1}{2}\right)^{n}$
Mass of radium left after $2$ half lives is
$m=m_{0} \left(\frac{1}{2}\right)^{n} =1\times\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=0.25\,mg$
Mass of radium disintegrated $=1-0.25=0.75\,mg $
Mass of radium disintegrated
$=0.75\times2.68\times10^{18}=2.01\times10^{18}$