• # question_answer A sample of radioactive material decays simultaneously by two processes A & B with half lives $\frac{1}{2}$ & $\frac{1}{4}hr$ respectively. For first half hr it decays with the process A, next one $hr$ with the process B & for further half an hour with both A & B. If originally there were ${{N}_{0}}$ nuclei the number of nuclei after 2 hr of such decay ${{N}_{0}}{{\left( \frac{1}{2} \right)}^{a}}$ find the value of a - A) 2                                  B) 4     C) 8          D) 1

Correct Answer: C

Solution :

[C]  For first half sample decay with process A of half lives $\frac{1}{2}hr.$ After first half hrs $N={{N}_{0}}\frac{1}{2}$ For next one hour decay process B of half lives $\frac{1}{4}$occur So for $t=\frac{1}{2}$ to $t=1\frac{1}{2}$ Total 4 half lives occur $N={{N}_{0}}\frac{1}{2}{{\left[ \frac{1}{2} \right]}^{4}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{5}}$
 For last $\frac{1}{2}hr$ both A and B process occur simultaneously For $t=1\frac{1}{2}$ to t = 2hr [For both A & B $\frac{1}{{{t}_{1/2}}}=\frac{1}{1/2}+\frac{1}{1/4}=6]$ ${{t}_{1/2}}=\frac{1}{6}hr$ Number of half-life in$\frac{1}{2}hr$ is 3 $\therefore \,\,\,N={{N}_{0}}{{\left[ \frac{1}{2} \right]}^{5}}\,\,{{\left[ \frac{1}{2} \right]}^{3}}$$={{N}_{0}}{{\left[ \frac{1}{2} \right]}^{8}}$ a = 8

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