Question

---

For an element decaying through simultaneous reaction, the half-life for the respective decaying path is $1400s$ and $700s$. Find the time taken when the number of atoms becomes$N_{\circ }/3N$ in the element sample $(N_{\circ }istheinitialnumberofatomsinsample)$.

• A. $(1400/5)\ln 3$
• B. $(1400/3)\ln 3$
• C. $(1400/3)\ln 2$
• D. $(700/3)\ln 2$

Answer 1

The correct option is B

$(1400/3)\ln 3$

Explanation for the correct options:

B) $\mathbf{(}\mathbf{1400}\mathbf{/}\mathbf{3}\mathbf{)}\mathit{l}\mathit{n}\mathbf{3}$

Step 1: the half-life for the decaying path is $t_{1=}1400,t_2=700s$

$Numberofatom\left(N\right)=3N_{\circ }$

Step 2: $Averagetime1/\tau =1/t_1+1/t_2$

$EquationforradioactiveintegrationN=N_{\circ }{e}^{-t/T}$

Step3: = $1/\tau =1/t_1+1/t_2$

$1/\tau =t_1{t}_2/t_1+t_2$

$$\begin{gathered} \tau =1400\times 700/2100 \\ \tau =1400/3 \end{gathered}$$

$N=N_{\circ }{e}^{-t/T}$

$N_{\circ }/3=N_{\circ }{e}^{-t/T}$

${\log }_e{e}^{t/\tau }={\log }_{e}3$=

$$\begin{gathered} t/\tau {\log }_e=\ln 3 \\ t=\tau \ln 3=1400/3\ln 3 \end{gathered}$$

The time taken by the $N_{\circ }/3N$atoms is $(1400/3)\ln 3$

As option B is correct, all the other options A, C and D options are incorrect.