Question

The half life of a radioactive substance is 20 minutes. In how much time the activity of substance drops to \((\frac{1}{16})^{th}\) of its initial value?

• A. 80 minutes
• B. 20 minutes
• C. 40 minutes
• D. 60 minutes

Answer 1

The Correct Option is A

To determine how much time it takes for the activity of a radioactive substance to drop to \((\frac{1}{16})^{th}\) of its initial value given a half-life of 20 minutes, we can use the formula for radioactive decay:

The formula for radioactive decay is:

\(N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{\frac{1}{2}}}}\)

Where:

• \(N(t)\) is the amount of substance remaining after time \(t\).
• \(N_0\) is the initial amount of the substance.
• \(T_{\frac{1}{2}}\) is the half-life of the substance.

We know that \(N(t) = \frac{1}{16} N_0\), and that the half-life \(T_{\frac{1}{2}} = 20\) minutes. We need to find the time \(t\).

Substituting this into the equation gives:

\(\frac{1}{16}N_0 = N_0 \left( \frac{1}{2} \right)^{\frac{t}{20}}\)

We can cancel \(N_0\) from both sides:

\(\frac{1}{16} = \left( \frac{1}{2} \right)^{\frac{t}{20}}\)

Converting the fraction \(\frac{1}{16}\) to base-2 gives us:

\(\frac{1}{16} = \left( \frac{1}{2} \right)^4\)

Thus, equating the exponents:

\(\frac{t}{20} = 4\)

Solving for \(t\), we multiply both sides by 20:

\(t = 4 \times 20 = 80\) minutes

Therefore, the time it takes for the activity to drop to \((\frac{1}{16})^{th}\) of its original value is 80 minutes. Hence, the correct answer is 80 minutes.