Question

Half-life of radium is $1600yr$. the sample of radium that would remain after 6400 year is

• A. $1/10$
• B. $1/16$
• C. $1/20$
• D. $1/4$

Answer 1

The correct option is B $1/16$

$\begin{array}{c}\text{Given,}\\ \text{half life of radium}=1600\text{year}\end{array}$

$\begin{array}{c}\text{Let intially no. of radium atom = No}\\ \text{we Know after any time (t) number become}\end{array}$

$N_t=n_0{e}^{-\lambda t}$

$\begin{array}{c}\text{Taking log both side}\\ \log \left(N_t\right)=\log N_0-\lambda t\\ \log \left(\frac{N_t}{N_0}\right)=-\lambda t\end{array}$

$\begin{array}{c}\text{we know}\\ \begin{array}{cc}{T}_{\frac{1}{2}}& =\frac{\log 2}{\lambda }\\ \lambda & =\frac{\log 2}{T_{1/2}}\end{array}\end{array}$

$\begin{array}{c}\text{Putting This value}\\ \begin{array}{cc}\log \left(\frac{N_t}{N_0}\right)& =\frac{-\log 2}{T_{1/2}}t\\ \frac{N_t}{N_0}& ={\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}\end{array}\end{array}$

$\begin{array}{cc}\text{Given}& t=6400\text{year}\\ & T_{1/2}=1600\text{year}\\ & \frac{N_t}{N_0}={\left(\frac{1}{2}\right)}^{\frac{6400}{1600}}\\ & \frac{N_t}{N_0}={\left(\frac{1}{2}\right)}^4=\frac{1}{16}\end{array}$

$\begin{array}{c}\text{fraction of sample that would remain}\\ \text{after}6400\text{year}=\frac{1}{16}\end{array}$