The numbers of nuclei of a radioactive substance at time t = 0 are 1000 and 900 at time t = 2 sec. Then the nu

Question

The numbers of nuclei of a radioactive substance at time t = 0 are 1000 and 900 at time t = 2 sec. Then the number of nuclei at time t = 4 sec will be:

Accepted Answer

1. **Principle of Radioactive Decay:** Radioactive decay follows first-order kinetics, which implies that the fraction of nuclei decaying in equal time intervals is constant .
2. **Calculate Decay Ratio:** 
   - At $t=0$, $N_0 = 1000$.
   - At $t=2$ s, $N_1 = 900$.
   - The fraction of nuclei remaining after 2 seconds is $\frac{N_1}{N_0} = \frac{900}{1000} = 0.9$.
3. **Apply to Next Interval:** 
   - The next time interval is also 2 seconds (from $t=2$ to $t=4$).
   - Therefore, the number of nuclei at $t=4$ ($N_2$) will be 0.9 times the number at $t=2$.
   - $N_2 = N_1 	imes 0.9 = 900 	imes 0.9 = 810$.
4. **Alternative Method (Formula):** Using $N = N_0 e^{-\lambda t}$, we have $900 = 1000 e^{-2\lambda} \Rightarrow e^{-2\lambda} = 0.9$. Then $N(4) = 1000 e^{-4\lambda} = 1000 (e^{-2\lambda})^2 = 1000 (0.9)^2 = 1000(0.81) = 810$.

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