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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Chemistry
Chapter
Structure of Atom

JIPMER 2016 · Chemistry · Structure of Atom

The effective neutron capture radius of a nucleus having a...

The effective neutron capture radius of a nucleus having a cross-section of 1.0 barr is [Given, 1 barr =1.0×10−24 cm2=1.0 \times 10^{-24} \mathrm{~cm}^2 ]

Options

  1. A5.6×10−13 cm5.6 \times 10^{-13} \mathrm{~cm}
  2. B4.3×10−13 cm4.3 \times 10^{-13} \mathrm{~cm}
  3. C2.3×10−11 cm2.3 \times 10^{-11} \mathrm{~cm}
  4. D5.6×10−24 cm5.6 \times 10^{-24} \mathrm{~cm}

Correct answer

A. 5.6×10−13 cm5.6 \times 10^{-13} \mathrm{~cm}

Step-by-step solution

1.0 barr =1.0×10−24 cm2=10−28 m2=1.0 \times 10^{-24} \mathrm{~cm}^2=10^{-28} \mathrm{~m}^2 The area of a circle is given by A=πr2A=\pi r^2 Hence, r=A/π=(1.0×10−24 cm2)/3.14=5.6×10−13 cm\begin{aligned} r & =\sqrt{A / \pi} \\ & =\sqrt{\left(1.0 \times 10^{-24} \mathrm{~cm}^2\right) / 3.14} \\ & =5.6 \times 10^{-13} \mathrm{~cm}\end{aligned}
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