Question

The activity of a radioactive substance is $4700$ per minute. Five minutes later the activity is $2700$ per minute. The decay constant is

• A. $0.1108mi{n}^{-1}$
• B. $1.1108mi{n}^{-1}$
• C. $2.008mi{n}^{-1}$
• D. $0.0108mi{n}^{-1}$

Answer 1

The correct option is A

$0.1108mi{n}^{-1}$

Step.1 Given data,

Initial counts per minute for radioactive substances,
$N_o=4700$
Final count, $N=2700$ and $t=5min$

Step. 2 Formula used,
$N=N_o{e}^{-\lambda t}$
The initial activity is $N_o$and the final activity is $N$, $\lambda$ is the decay constant.

Step. 3 Calculating the decay constant,
$N_o=4700$
$2700$ counts per minute are the activity after $5$ minutes,
$N=2700$ and $t=5min$
The first order decay equation is,
$N=N_o{e}^{-\lambda t}$
Substituting the values we get,
$2700=4700e^{\left(-\lambda \times 5\right)}$
${\displaystyle \frac{2700}{4700}}=e^{\left(-\lambda \times 5\right)}$
$-5\lambda =\ln \left({\displaystyle \frac{27}{47}}\right)$
$-5\lambda =-0.55$
Therefore, the value of the decay constant as,
$\lambda =0.1108\stackrel{-1}{min}$

Hence option A is the correct answer.