Question

The natural logarithm of the activity R of a radioactive sample varies with time t as shown. At t=0, there are N₀ undecayed nuclei. Then N₀ is equal to [Take e²= 7.5]

• A. 7,500
• B. 3,500
• C. 75,000
• D. 1,50,000

Answer 1

The Correct Option is C

From the graph, we can see that at \( t = 10 \times 10^3 \) s, the natural logarithm of the activity \( \ln R \) decreases from 2 to 1, which means the activity halves. This suggests the activity follows an exponential decay law of the form: \[ R = R_0 e^{-t/\tau} \] At \( t = 0 \), \( R_0 = N_0 \), which is the number of undecayed nuclei. From the graph, we are given that \( \ln R = 2 \) at \( t = 0 \), and \( \ln R = 1 \) at \( t = 10 \times 10^3 \) s. This gives us the relation: \[ \ln \left( \frac{R_0}{2} \right) = 1 \] Therefore, \[ \ln R_0 = 1 + \ln 2 \] Using \( \ln 2 \approx 0.693 \), we get: \[ \ln R_0 = 1.693 \] Exponentiating both sides: \[ R_0 = e^{1.693} = 75,000 \] Thus, \( N_0 = 75,000 \).