Q107 / 120 · VITEEE 2012 · Physics

Highly energetic electrons are bombarded on a target of an element contain ing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is $(14)^{1 / 3} .$ The atomic number of nucleus will be

  • A
    25
  • Correct answer
    B
    26
  • C
    56
  • D
    30

Solution

$$ \begin{array}{l} \text { Using } R=R_{0} A^{1 / 3} \\ \frac{R_{1}}{R_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{1 / 3} \\ \frac{R}{R_{\text {He }}}=\left(\frac{A}{4}\right)^{\frac{1}{3}} \end{array} $$ $$ (14)^{1 / 3}=\left(\frac{\mathrm{A}}{4}\right)^{\frac{1}{3}} \Rightarrow \mathrm{A}=56 $$ So, $=56-30=26$

Hence, the correct answer is option: 'B'.

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