Question

The activity of a radioactive substance falls from 700 s^-1 to 500 s^-1 in 30 minutes. Its half-life is close to

• A. $66min$
• B. $62min$
• C. $52min$
• D. $72min$

Answer 1

The correct option is B

$62min$

Step 1: Given data

The initial activity of radioactive substance $A_o$=$700s^{-1}$

The activity after $30min$ $A_t$=$500s^{-1}$

Step 2: Formula Used

It is understood that the radioactive decay is computed using the formula $A_t=A_{\circ }{e}^{-\lambda t}$

Where $A_t=Amountofsubs\tan ceattimet$, $A_{\circ }=Initialamount$ and $\lambda =decaycons\tan t$

Step 3: Calculating the half-life

From the formula,

$$\begin{gathered} A_t=A_{\circ }{e}^{-\lambda t} \\ \frac{A_{\circ }}{A_t}=e^{\lambda t} \\ \ln \left(\frac{A_{\circ }}{A_t}\right)=\lambda t----\left(1\right) \end{gathered}$$

We know that for half-life,

$A_t=\frac{A_{\circ }}{2}\&t=t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

So,

$$\begin{gathered} \ln \left(\frac{A_{\circ }}{{\displaystyle \raisebox{1ex}{$A_{\circ }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)=\lambda t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \ln \left(2\right)=\lambda t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}----\left(2\right) \end{gathered}$$

According to the given condition, substitute the known value in the equation $\left(1\right)$,

$\ln \left(\frac{700}{500}\right)=\lambda \times 30----\left(3\right)$

Divide equation $\left(2\right)$ by equation $\left(3\right)$,

$$\begin{gathered} \frac{\ln \left(2\right)}{\ln \left({\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\right)}=\frac{t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{30} \\ {t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln \left(2\right)\times 30}{\ln \left(\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)} \\ {t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0·693\times 30}{0·336} \\ {t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=61·8min \end{gathered}$$

So, the half-life of a radioactive substance is approximately $62min$.

Hence, option B is the correct answer.