Question

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

• A. $2.01\times 10^9$
• B. $2.01\times 10^{18}$
• C. $1.01\times 10^9$
• D. $1.01\times 10^{18}$

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
$m=m_0{\left(\frac{1}{2}\right)}^n=1\times {\left(\frac{1}{2}\right)}^2=\frac{1}{4}=0.25mg$
Mass of radium disintegrated $=1-0.25=0.75mg$
Mass of radium disintegrated
$=0.75\times 2.68\times 10^{18}=2.01\times 10^{18}$