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QAIPMT-2010-Ph-1988: NEET 2010 Physics - Class 12 AIPMT-2010-Ph-1988 Solution: The decay constant of a radio isotope is \( \). If \(A_1\) and…

The decay constant of a radio isotope is λ\lambda. If A1A_1 and A2A_2 are its activities at times t1t_1 and t2t_2 respectively, the number of nuclei which have decayed during the time (t1−t2)(t_1-t_2) is

Multiple Choice Options

  • A A1t1−A2t2A_1t_1-A_2t_2
  • B A1−A2A_1-A_2
  • C (A1−A2)/λ(A_1-A_2)/\lambda
  • D λ(A1−A2)\lambda(A_1-A_2)

Step-by-Step Verified Solution

A 1 = _ N 1 at time t 1 A 2 = _ N 2 at time t 2 Therefore, number of nuclei decayed during time interval (t 1 � t 2 ) is N 1 - N 2 = [ A 1 _ A 2 ] _

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