Question

A radioactive sample has half-life of 3 years. The time required for the activity of the sample to reduce to \(\frac{1}{5}\)th of its initial value is about

• A. 7 years
• B. 15 years
• C. 5 years
• D. 10 years

Answer 1

The Correct Option is A

Given: 

• Half-life of the radioactive sample, \( T = 3 \) years
• Final activity is \( \frac{1}{5} \) of the initial activity

Decay Formula:

\[ N = N_0 \times \left( \frac{1}{2} \right)^{t/T} \]

Substituting values:

\[ \frac{1}{5} = \left( \frac{1}{2} \right)^{t/3} \]

Taking natural logarithm on both sides:

\[ \ln \left( \frac{1}{5} \right) = \frac{t}{3} \ln \left( \frac{1}{2} \right) \]

Using values:

\[ \ln(1/5) = -\ln 5 \approx -1.609 \]

\[ \ln(1/2) = -\ln 2 \approx -0.693 \]

Solving for \( t \):

\[ t = \frac{1.609 \times 3}{0.693} \]

\[ t \approx \frac{4.827}{0.693} \approx 7 \text{ years} \]

Answer: The time required is 7 years (Option A).

Answer 2

The activity of a radioactive sample follows the exponential decay formula: \[ A = A_0 e^{-\lambda t} \] where \( A_0 \) is the initial activity, \( A \) is the activity after time \( t \), and \( \lambda \) is the decay constant. The half-life \( T_{1/2} \) is related to the decay constant by the formula: \[ T_{1/2} = \frac{\ln 2}{\lambda} \] Given that the half-life is 3 years, we can calculate \( \lambda \): \[ \lambda = \frac{\ln 2}{3} \approx 0.231 \] Now, we want to find the time required for the activity to reduce to \( \frac{1}{5} \) of its initial value. Substituting into the exponential decay equation: \[ \frac{A}{A_0} = \frac{1}{5} = e^{-\lambda t} \] Taking the natural logarithm of both sides: \[ \ln \left( \frac{1}{5} \right) = -\lambda t \] \[ t = \frac{\ln 5}{\lambda} \approx \frac{1.609}{0.231} \approx 6.96 \, \text{years} \] Thus, the correct answer is \((A) \, 7 \, \text{years}\).