Question

A sample originally contained ${10}^{20}$radioactive atoms, which emit $\alpha$ -particles emitted inthe thirdyear to that emitted during the second year is $0.3$.How many $\alpha$ particles were emitted in the first year?

• A. $7\times {10}^{19}$
• B. $3\times {10}^{19}$
• C. $5\times {10}^{18}$
• D. $3\times {10}^{18}$

Answer 1

The correct option is A $7\times {10}^{19}$

$\begin{array}{c}\text{Let}{N}_0\text{be initial amount of atom which is}{10}^{20}\\ \text{and say}\lambda \text{is decay constant}\end{array}$

$\begin{array}{c}\text{we know}\\ N_t=N_0{e}^{-\lambda t},N_t=\text{No of atom left after t}\end{array}$

$\begin{array}{c}\text{after first year.}\\ N_1=N_0{e}^{-\lambda }\end{array}$

$\begin{array}{c}\text{So, amount decayed in first year will be}\\ N_0-N_0{e}^{-\lambda }\\ \text{No}(1-e^{-\lambda })\end{array}$

$\begin{array}{cc}\text{given}& \\ \frac{N_2}{N_3}=0.3& \\ \text{Or}{N}_2& =N_0{e}^{-2\lambda }\\ N_3& =N_0{e}^{-3\lambda }\end{array}$

$\begin{array}{c}\frac{N_0{e}^{-3\lambda }}{N_0{e}^{-2\lambda }}=0.3\\ e^{-\lambda }=0.3\end{array}$

$\text{Now equation (1) become}$

$\begin{array}{cc}& \text{No}(1-0.3)\\ =& 0.7\times {10}^{20}\because N_0={10}^{20}\\ =& 7\times {10}^{19}\text{atom}\end{array}$

$\text{So, amount decayed in}{1}^{\text{st}}\text{year}=7\times {10}^{19}$