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AIIMS 201926 May
Exam
AIIMS
Year
2019
Date
26 May
Subject
Physics
Chapter
Nuclear Physics

AIIMS 2019 · 26 May · Physics · Nuclear Physics

If half life of an element is 69.3 hours then how much of its...

If half life of an element is 69.3 hours then how much of its percent will decay between 10th 10^{\text{th }} and 11th 11^{\text{th }} hour. Initial activity =50μCi.=50 \mu \mathrm{C}_i .

Options

  1. A1%1 \%
  2. B2%2 \%
  3. C3%3 \%
  4. D4%4 \%

Correct answer

A. 1%1 \%

Step-by-step solution

Let active Nuclei at t=0t=0 is N0N_0 then, active nuclei at t=10t=10 hour N1=N0e−10λN_1=N_0 e^{-10 \lambda} Active nuclei at t=11t=11 hour N2=N0e−11λN_2=N_0 e^{-11 \lambda}% decay =(N1−N2N1)×100=(N0⋅e−10λ−N0e−11λN1)×100=(N0⋅e−10λ−N0e−11λN0⋅e−10λ)×100=(1−1eλ)×100=(1−1eln⁡2/T1/2)×100=(∵T1/2=69.3hrs.)=(1−1e0.01)×100⇒% decay =1%.\begin{aligned} & \% \text{ decay }=\left(\frac{N_1-N_2}{N_1}\right) \times 100 \\ & =\left(\frac{N_0 \cdot e^{-10 \lambda}-N_0 e^{-11 \lambda}}{N_1}\right) \times 100 \\ & =\left(\frac{N_0 \cdot e^{-10 \lambda}-N_0 e^{-11 \lambda}}{N_0 \cdot e^{-10 \lambda}}\right) \times 100 \\ & =\left(1-\frac{1}{e^\lambda}\right) \times 100=\left(1-\frac{1}{e^{\ln 2 / T_{1/2}}}\right) \times 100 \\ & =\left(\because T_{1/2}=69.3 \mathrm{hrs} .\right) \\ & =\left(1-\frac{1}{e^{0.01}}\right) \times 100 \Rightarrow \% \text{ decay }=1 \% .\end{aligned}$$
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