Question

\( 75\% \) of \( ^{234}_{90}\text{Th} \) decays in \( t \) years. Its half-life is (in years):

• A. \( t \)
• B. \( \dfrac{t}{2} \)
• C. \( 2t \)
• D. \( 4t \)
• E. \( \dfrac{t}{4} \)

Hint:

If remaining fraction becomes \( \frac{1}{2^n} \), then \( n \) half-lives have passed.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem deals with radioactive decay and the concept of half-life. The half-life (\(T_{1/2}\)) is the time required for half of the radioactive nuclei in a sample to decay. We are given the time it takes for 75% of a sample to decay and are asked to find the half-life.
Step 2: Key Formula or Approach:
Method 1: Using the decay formula
The number of undecayed nuclei N at time t is given by:
\[ N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} \] where \(N_0\) is the initial number of nuclei.
Method 2: Conceptual Approach
We can think about the decay process in terms of half-lives.
- After one half-life, 50% of the sample has decayed, and 50% remains.
- After a second half-life, half of the remaining 50% decays (which is 25% of the original), leaving 25% of the original sample.
Step 3: Detailed Explanation:
Method 2 (Conceptual Approach) is faster:
The problem states that 75% of the sample decays. This means that \( 100% - 75% = 25% \) of the sample remains.
Let's trace the remaining amount:
Start with 100% of the sample.
After 1 half-life (\(T_{1/2}\)), the remaining amount is \( \frac{1}{2} \times 100% = 50% \).
After another half-life (a total of 2 half-lives), the remaining amount is \( \frac{1}{2} \times 50% = 25% \).
So, for 75% of the sample to decay (leaving 25%), a time equivalent to two half-lives must have passed.
We are given that this time is \( t \) years.
Therefore, \( t = 2 \times T_{1/2} \).
We need to find the half-life, \(T_{1/2}\). Rearranging the equation:
\[ T_{1/2} = \frac{t}{2} \] Method 1 (Formulaic Approach):
If 75% has decayed, the remaining fraction is \( \frac{N(t)}{N_0} = 1 - 0.75 = 0.25 = \frac{1}{4} \).
Using the decay formula:
\[ \frac{1}{4} = \left(\frac{1}{2}\right)^{t/T_{1/2}} \] Since \( \frac{1}{4} = \left(\frac{1}{2}\right)^2 \), we can write:
\[ \left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{t/T_{1/2}} \] By comparing the exponents:
\[ 2 = \frac{t}{T_{1/2}} \] \[ T_{1/2} = \frac{t}{2} \] Step 4: Final Answer:
The half-life of the substance is \( \frac{t}{2} \) years. This corresponds to option (B).