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NEET PHYSICSNUCLEIEasy

Question

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 .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from NUCLEI. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSNUCLEIonesixteenthstartingamountcertainradioactive

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