Question

Consider two radionuclides P and Q. Suppose the half-life of P (\( t_{1/2}^P \)) is four times that of Q (\( t_{1/2}^Q \)). At time \( t = 0 \), there are \( N_0 \) atoms of both radionuclides. When will the radioactivity of the two radionuclides be equal?

• A. \( t = t_{1/2}^P \)
• B. \( t = 0.66\, t_{1/2}^P \)
• C. \( t = 0.75\, t_{1/2}^P \)
• D. \( t = 1.5\, t_{1/2}^P \)

Hint:

Activity equality problems often reduce to comparing decay constants; use the relationship \( t_{1/2} \propto 1/\lambda \).

Answer 1

The Correct Option is B

Step 1: Write decay constants.
Decay constant is related to half-life as:
\[ \lambda = \frac{\ln 2}{t_{1/2}} \]
Given \( t_{1/2}^P = 4 t_{1/2}^Q \), we get:
\[ \lambda_P = \frac{\ln 2}{4 t_{1/2}^Q}, \quad \lambda_Q = \frac{\ln 2}{t_{1/2}^Q} \]
Thus,
\[ \lambda_Q = 4\lambda_P \]
Step 2: Write radioactivity expressions.
Radioactivity:
\[ A(t) = \lambda N(t) \]
\[ N(t) = N_0 e^{-\lambda t} \]
So radioactivities are:
\[ A_P(t) = \lambda_P N_0 e^{-\lambda_P t} \]
\[ A_Q(t) = \lambda_Q N_0 e^{-\lambda_Q t} \]
Step 3: Set the activities equal.
\[ \lambda_P e^{-\lambda_P t} = \lambda_Q e^{-\lambda_Q t} \]
\[ e^{-(\lambda_Q - \lambda_P)t} = \frac{\lambda_P}{\lambda_Q} \]
\[ e^{-3\lambda_P t} = \frac{1}{4} \]
Step 4: Take natural logarithm.
\[ -3\lambda_P t = -\ln 4 \]
\[ t = \frac{\ln 4}{3\lambda_P} \]
Step 5: Substitute \( \lambda_P = \frac{\ln 2}{t_{1/2}^P} \).
\[ t = \frac{\ln 4}{3} \cdot \frac{t_{1/2}^P}{\ln 2} \]
Since \( \ln 4 = 2\ln 2 \):
\[ t = \frac{2\ln 2}{3\ln 2} \, t_{1/2}^P \]
\[ t = \frac{2}{3} t_{1/2}^P \]
\[ t \approx 0.66\, t_{1/2}^P \]
Step 6: Conclusion.
Thus, the radioactivity of P and Q becomes equal at
\[ t = 0.66\, t_{1/2}^P \]
Hence, the correct answer is (B).