Question

A radioactive element undergoes 80% radioactive decay in 300 min. The half-life for this species in minutes is ............

Hint:

If 80% decays, only 20% remains → use N/N$_0$ = 0.20.

Answer 1

Correct Answer: 128

Step 1: Understand the decay information.
80% decays → 20% remains. So fraction remaining = 0.20.
Step 2: Use first-order decay law.
\[ N = N_0 e^{-kt} \] \[ 0.20 = e^{-k(300)} \] Step 3: Take natural logarithm.
\[ \ln(0.20) = -300k \] \[ k = \frac{-\ln(0.20)}{300} \] Step 4: Use half-life formula.
\[ t_{1/2} = \frac{0.693}{k} \] Step 5: Compute numerically.
\[ k = \frac{1.609}{300} = 0.00536\text{ min}^{-1} \] \[ t_{1/2} = \frac{0.693}{0.00536} \approx 129.4 \approx 134 \text{ min} \] Step 6: Conclusion.
Thus, the half-life is approximately 134 minutes.