A sample of a radioactive isotope has an initial activity of 8000\text{ Bq}. After an elap… — JAMB UTME

A sample of a radioactive isotope has an initial activity of $8000\text{ Bq}$ . After an elapsed time of $18\text{ minutes}$ , its activity reduces to $1000\text{ Bq}$ . Calculate the decay constant $\lambda$ of the isotope in $\text{min}^{-1}$ .

Answer: 0.1155 min^-1

Answer

The decay constant of the radioactive isotope is

0.1155\text{ min}^{-1}

The activity decreases from

8000\text{ Bq}

1000\text{ Bq}

, which is a reduction to

\frac{1}{8}

of its initial value. Since

\left(\frac{1}{2}\right)^3 = \frac{1}{8}

, exactly

3

half-lives have elapsed in

18\text{ minutes}

, meaning

T_{1/2} = 6\text{ minutes}

. Using the relationship

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{6\text{ min}}

, we obtain

\lambda \approx 0.1155\text{ min}^{-1}

Step-by-Step Solution

Determine the number of elapsed half-lives from the activity reduction ratio.

The fraction of remaining activity is

\frac{A}{A_0} = \frac{1000}{8000} = \frac{1}{8} = \left(\frac{1}{2}\right)^3

, giving

n = 3

half-lives.

Radioactive decay follows the relation

A = A_0 (1/2)^n

Determine the half-life

T_{1/2}

of the isotope.

T_{1/2} = \frac{t}{n} = \frac{18\text{ min}}{3} = 6\text{ minutes}

Total elapsed time is equal to the number of half-lives multiplied by the duration of one half-life.

Compute the decay constant

\lambda

\text{min}^{-1}

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{6\text{ min}} = 0.1155\text{ min}^{-1}

The decay constant is fundamental to decay rate and related to half-life via

\lambda = \frac{\ln 2}{T_{1/2}}

Key Concept

Radioactive Decay Law and Decay Constant

Estimated Time:2m 0s

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