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, \( N_0 = 75,000 \). 

Answer 2

The activity \( R \) of a radioactive sample is related to the number of undecayed nuclei \( N \) by the equation: \[ R = \lambda N \] where \( \lambda \) is the decay constant and \( N \) is the number of undecayed nuclei at time \( t \). The relationship between the activity and the number of undecayed nuclei can also be written as: \[ \ln R = \ln R_0 - \lambda t \] where \( R_0 \) is the initial activity at \( t = 0 \) and \( \lambda \) is the decay constant. From the given graph, we observe that \( \ln R \) is plotted against time \( t \), and the curve suggests an exponential decay. The initial value of \( \ln R \) at \( t = 0 \) corresponds to the natural logarithm of the initial activity, \( \ln R_0 \). The graph shows that when \( t = 10 \times 10^3 \, \text{s} \), \( \ln R = 1 \). This gives us the equation: \[ \ln R_0 - \lambda (10 \times 10^3) = 1 \] Additionally, at \( t = 0 \), the natural logarithm of the initial activity is \( \ln R_0 = 2 \), so: \[ \ln R_0 = 2 \] Thus, we have: \[ 2 - \lambda (10 \times 10^3) = 1 \] Solving for \( \lambda \): \[ \lambda (10 \times 10^3) = 1 \quad \Rightarrow \quad \lambda = \frac{1}{10 \times 10^3} = 10^{-4} \, \text{s}^{-1} \] Now, using the equation for radioactive decay: \[ N = N_0 e^{-\lambda t} \] At \( t = 0 \), we know that \( N_0 \) is the number of undecayed nuclei, and the activity \( R_0 \) is related to \( N_0 \) by \( R_0 = \lambda N_0 \). Since \( \ln R_0 = 2 \), we have: \[ R_0 = e^2 = 7.5 \] Thus, \( R_0 = \lambda N_0 \), and: \[ 7.5 = 10^{-4} \times N_0 \] Solving for \( N_0 \): \[ N_0 = \frac{7.5}{10^{-4}} = 75,000 \] Therefore, the initial number of undecayed nuclei is \( N_0 = 75,000 \).