Question

If half life for a radioactive decay reaction is T. Find the time after which \(\frac{7}{8}th\) of initial mass decays

• A. 3T
• B. 2T
• C. \(\frac{T}{2}\)
• D. 4T

Answer 1

The Correct Option is A

To solve this question, we need to understand the concept of half-life in radioactive decay. The half-life \( T \) is the time required for a substance to reduce to half its initial amount.

We are given that the half-life of the substance is \( T \) and we need to find the time after which \(\frac{7}{8}\) of the initial mass decays.

This implies that only \(\frac{1}{8}\) of the original mass remains. The fundamental formula for decay in terms of the remaining fraction is:

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

Here, \( N_0 \) is the initial quantity, \( N \) is the remaining quantity, \( t \) is the time elapsed, and \( T \) is the half-life.

To find when \(\frac{1}{8}\) of the substance is left, set \( N = \frac{N_0}{8} \):

\(\frac{N_0}{8} = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T}}\)

Divide both sides by \( N_0 \) and simplify:

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

Express \(\frac{1}{8}\) as a power of \(\frac{1}{2}\):

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

From this, comparing the exponents, we have:

\(\frac{t}{T} = 3\)

Therefore, \( t = 3T \).

Thus, the correct answer is 3T. The time after which \(\frac{7}{8}\) of the initial mass decays is 3 times the half-life of the substance.