Question

The half-life of \(^{113m}In\) nuclei is \(1.7\) hours, and the \(^{99m}Tc\) nuclei is \(6.0\) hours. If the initial activity levels of \(^{113m}In\) and \(^{99m}Tc\) in their respective samples are \(100\) Gigabecquerels (GBq) and \(20\) GBq, the time at which their activity levels become equal is hours. (Round off to one decimal place)

Hint:

Write each activity as \(A_0 e^{-\lambda t}\) with \(\lambda = \ln2/T_{1/2}\), set the two equal, and solve for \(t\).

Answer 1

Correct Answer: 5.5

Step 1: Write the decay law for each isotope.
Radioactive decay follows \(A(t) = A_0 e^{-\lambda t}\), where \(A_0\) is the initial activity and \(\lambda\) is the decay constant, related to the half-life \(T_{1/2}\) by \(\lambda = \dfrac{\ln 2}{T_{1/2}}\).
For \(^{113m}In\): \(A_0 = 100\) GBq, \(T_{1/2} = 1.7\) h, so \(\lambda_1 = \dfrac{\ln 2}{1.7} = 0.4077\ \text{h}^{-1}\).
For \(^{99m}Tc\): \(A_0 = 20\) GBq, \(T_{1/2} = 6.0\) h, so \(\lambda_2 = \dfrac{\ln 2}{6.0} = 0.1155\ \text{h}^{-1}\).

Step 2: Set the two activities equal.
The activities become equal when:
\[ 100\, e^{-\lambda_1 t} = 20\, e^{-\lambda_2 t} \]
\[ \frac{100}{20} = e^{-\lambda_2 t + \lambda_1 t} = e^{(\lambda_1 - \lambda_2)t} \]
\[ 5 = e^{(\lambda_1 - \lambda_2)t} \]

Step 3: Take the natural log and solve for t.
\[ \ln 5 = (\lambda_1 - \lambda_2)\, t \]
\[ \lambda_1 - \lambda_2 = 0.4077 - 0.1155 = 0.2922\ \text{h}^{-1} \]
\[ t = \frac{\ln 5}{0.2922} = \frac{1.6094}{0.2922} \approx 5.51\ \text{h} \]

Step 4: Check the physical sense.
\(^{113m}In\) starts with a much higher activity but decays faster (shorter half-life), while \(^{99m}Tc\) starts lower but decays more slowly. So the faster-decaying \(^{113m}In\) activity falls and eventually meets the slower-decaying \(^{99m}Tc\) activity at around \(t \approx 5.5\) h. Rounded to one decimal place, \(t \approx 5.5\) hours.

Final Answer:
The activity levels become equal at \(t \approx 5.5\) hours. \[ \boxed{t \approx 5.5\ \text{hours}} \]