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NEET PHYSICSEasy

After two hours, one-sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half life of the isotope is

A

15 minutes

B

30 minutes

C

45 minutes

D

1 hour

Step-by-Step Solution

  1. Radioactive Decay Law: Radioactive decay follows first-order kinetics. The amount of substance remaining (NN) is related to the initial amount (N0N_0) and the number of half-lives (nn) by the formula: NN0=(12)n\frac{N}{N_0} = \left(\frac{1}{2}\right)^n .
  2. Determine Number of Half-Lives (nn):
  • Given fraction remaining NN0=116\frac{N}{N_0} = \frac{1}{16}.
  • We know that 116=(12)4\frac{1}{16} = \left(\frac{1}{2}\right)^4.
  • Therefore, the number of half-lives passed is n=4n = 4.
  1. Calculate Half-Life (T1/2T_{1/2}):
  • Total time elapsed (tt) = 2 hours = 120 minutes.
  • The relationship between total time and half-life is t=n×T1/2t = n \times T_{1/2}.
  • 120 min=4×T1/2120 \text{ min} = 4 \times T_{1/2}.
  • T1/2=1204=30T_{1/2} = \frac{120}{4} = 30 minutes .
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