Question:
The natural logarithm of the activity \( R \) of a radioactive sample varies with time \( t \) as shown. At \( t = 0 \), there are \( N_0 \) undecayed nuclei. Then \( N_0 \) is equal to \([\text{Take} \ e^{7.5}] \)
The natural logarithm of the activity \( R \) of a radioactive sample varies with time \( t \) as shown. At \( t = 0 \), there are \( N_0 \) undecayed nuclei. Then \( N_0 \) is equal to \([\text{Take} \ e^{7.5}] \)
Updated On: Dec 26, 2024
- 7,500
- 3,500
- 75,000
- 1,50,000
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The Correct Option is C
Solution and Explanation
From the given graph, we find the initial activity \( R_0 \) at \( t = 0 \) and its natural logarithm:
\(\ \ \log_e R_0 = 7.5 \)
Thus,\(\ R_0 = e^{7.5} \approx 75,000\)
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