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NEET PHYSICSEasy

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:

A

800

B

810

C

790

D

700

Step-by-Step Solution

  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=0t=0, N0=1000N_0 = 1000.
  • At t=2t=2 s, N1=900N_1 = 900.
  • The fraction of nuclei remaining after 2 seconds is N1N0=9001000=0.9\frac{N_1}{N_0} = \frac{900}{1000} = 0.9.
  1. Apply to Next Interval:
  • The next time interval is also 2 seconds (from t=2t=2 to t=4t=4).
  • Therefore, the number of nuclei at t=4t=4 (N2N_2) will be 0.9 times the number at t=2t=2.
  • N2=N1×0.9=900×0.9=810N_2 = N_1 \times 0.9 = 900 \times 0.9 = 810.
  1. Alternative Method (Formula): Using N=N0e−λtN = N_0 e^{-\lambda t}, we have 900=1000e−2λ⇒e−2λ=0.9900 = 1000 e^{-2\lambda} \Rightarrow e^{-2\lambda} = 0.9. Then N(4)=1000e−4λ=1000(e−2λ)2=1000(0.9)2=1000(0.81)=810N(4) = 1000 e^{-4\lambda} = 1000 (e^{-2\lambda})^2 = 1000 (0.9)^2 = 1000(0.81) = 810.
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