Question

An $\alpha$-particle of energy $5 , \mathrm{MeV}$ is scattered through $180^\circ$ by a gold nucleus. The distance of closest approach is of the order of

A. $10^{-12} , \mathrm{cm}$

B. $10^{-16} , \mathrm{cm}$

C. $10^{-10} , \mathrm{cm}$

D. $10^{-14} , \mathrm{cm}$

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Answer

:

To determine the distance of closest approach for an $\alpha$-particle of energy $5 ,\mathrm{MeV}$ scattered by a gold nucleus through $180^\circ$, consider the conservation of energy principle.

  1. Initial kinetic energy (K.E.) of the $\alpha$-particle converts to electrostatic potential energy at the point of closest approach: $$ \frac{1}{2}mv^2 = \frac{K(2e)(79e)}{r} $$ where $m$ is the mass, $v$ is the velocity, $K$ is Coulomb's constant, and $r$ is the distance of closest approach.

  2. Given: $$ \frac{1}{2}mv^2 = 5 , \text{MeV} = 5 \times 10^6 , \text{eV} $$

  3. Solving for $r$: $$ r = \frac{K(2e)(79e)}{5 \times 10^6 , \text{eV}} $$

  4. Substituting the values: $$ K = 9 \times 10^9 , \text{Nm}^2\text{C}^{-2} $$ $e = 1.6 \times 10^{-19} , \text{C} $$ This gives: $$ r = \frac{9 \times 10^9 \times 2 \times 1.6 \times 10^{-19} \times 79 \times 1.6 \times 10^{-19}}{5 \times 10^6 \times 1.6 \times 10^{-13}} $$

  5. Simplifying: $$ r \approx 10^{-12} , \text{cm} $$

Final Answer: A


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