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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Nuclear Physics

JIPMER 2016 · Physics · Nuclear Physics

A^7 Li target is bombarded with a proton beam current of 10^ -4 ~A...

A7Li\mathrm{A}^7 \mathrm{Li} target is bombarded with a proton beam current of 10−4 A10^{-4} \mathrm{~A} for one hour to produce 7Be{ }^7 \mathrm{Be} of activity 1.8×1081.8 \times 10^8 disintegrations per second. Assuming that one 7{ }^7 Be radioactive nuclei is produced by bombarding 1000 protons, its half-life isA^7 Li target is bombarded with a proton beam current of 10^ -4 ~A for one hour to produce ^7 Be of activity 1.8 × 10^8

Options

  1. A0.87×107 s0.87 \times 10^7 \mathrm{~s}
  2. B0.2×107 s0.2 \times 10^7 \mathrm{~s}
  3. C0.67×108 s0.67 \times 10^8 \mathrm{~s}
  4. D0.87×106 s0.87 \times 10^6 \mathrm{~s}

Correct answer

A. 0.87×107 s0.87 \times 10^7 \mathrm{~s}

Step-by-step solution

A^7 Li target is bombarded with a proton beam current of 10^ -4 ~A for one hour to produce ^7 Be of activity 1.8 × 10^8 Given, i=10−4 A,t=1 h=3600 si=10^{-4} \mathrm{~A}, t=1 \mathrm{~h}=3600 \mathrm{~s} We know that, q=q= it =10−4×1×3600=0.36C∴=10^{-4} \times 1 \times 3600=0.36 \mathrm{C} \therefore Number of protons =qe=0.361.6×10−19[∵e=1.6×10−19C]=2.25×108=\frac{q}{e}=\frac{0.36}{1.6 \times 10^{-19}} \left[\because e=1.6 \times 10^{-19} \mathrm{C}\right] =2.25 \times 10^8 Number of 7{ }^7 Be nuclei produced =2.25×10181000=2.25×1015=\frac{2.25 \times 10^{18}}{1000}=2.25 \times 10^{15} Activity of 7Be=1.8×108{ }^7 \mathrm{Be}=1.8 \times 10^8 disintegration /s/ \mathrm{s}∴ Activity, A=λN=ln⁡2t1/2N\begin{aligned} \therefore \text{ Activity, } A & =\lambda N \\ & =\frac{\ln 2}{t_{1/2}} N\end{aligned}[∵λ=ln⁡2t1/2]1.8×108=0.693t1/2×2.25×1015∴t1/2=0.693×2.25×10151.8×108=0.87×107 s\left[\because \lambda=\frac{\ln 2}{t_{1/2}}\right] 1.8 \times 10^8=\frac{0.693}{t_{1/2}} \times 2.25 \times 10^{15} \therefore \quad t_{1/2}=\frac{0.693 \times 2.25 \times 10^{15}}{1.8 \times 10^8}=0.87 \times 10^7 \mathrm{~s}
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