Q137 / 200 · NEET 2010 (Screening) · Physics

An alpha nucleus of energy $\frac{1}{2} \mathrm{mv}^2$ bombards a heavy nuclear target of charge $Z e$. Then the distance of closest approach for the alpha nucleus will be proportional to

  • A
    $\frac{1}{\mathrm{Ze}}$
  • B
    $v^2$
  • Correct answer
    C
    $\frac{1}{\mathrm{~m}}$
  • D
    $\frac{1}{\mathrm{v}^4}$

Solution

An $\alpha$-particle of mass $m$ possesses initial velocity $\mathrm{v}$, when it is at a large distance from the nucleus of an atom having atomic number $Z$. At the distance of closest approach, the kinetic energy of $\alpha$-particle is completely converted into potential energy. Mathematically, $$ \begin{aligned} & \frac{1}{2} \mathrm{mv}^2=\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}_0} \\ & \mathrm{r}_0=\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{Ze}^2}{\frac{1}{2} \mathrm{mv}^2} \end{aligned} $$

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

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