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: Interpret the given statement.
It is given that \( 75\% \) of the radioactive substance decays in time \( t \).
This means only \( 25\% \) of the substance remains: \[ N = 0.25 N_0 = \frac{N_0}{4} \]

Step 2: Use radioactive decay law.
\[ N = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} \] where \( T_{1/2} \) is the half-life.

Step 3: Substitute the remaining fraction.
\[ \frac{N_0}{4} = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} \]

Step 4: Simplify the equation.
\[ \frac{1}{4} = \left(\frac{1}{2}\right)^{t/T_{1/2}} \] But, \[ \frac{1}{4} = \left(\frac{1}{2}\right)^2 \]

Step 5: Compare powers.
\[ \left(\frac{1}{2}\right)^{t/T_{1/2}} = \left(\frac{1}{2}\right)^2 \Rightarrow \frac{t}{T_{1/2}} = 2 \]

Step 6: Solve for half-life.
\[ T_{1/2} = \frac{t}{2} \]

Step 7: Final conclusion.
Hence, the half-life is: \[ \boxed{\frac{t}{2}} \] Therefore, the correct option is \[ \boxed{(2)\ \dfrac{t}{2}} \]