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(B) In the Rutherford $\alpha - $particle scattering experiment,the scattering of $\alpha - $particles by the gold foil was explained by assuming that

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The force of repulsion between an atomic nucleus and an $\alpha - $ particle varies with distance according to inverse square law.

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(B) In the Rutherford $\alpha - $particle scattering experiment,the scattering of $\alpha - $particles by the gold foil was explained by assuming that the electrostatic force of repulsion between the positively charged nucleus and the positively charged $\alpha - $particle follows Coulomb's law.Since Coulomb's law states that the force $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$,the force is inversely proportional to the square of the distance $(F \propto \frac{1}{r^2})$.This inverse square law dependence was essential for Rutherford to derive the scattering formula that matched the experimental observations.

Which one did Rutherford consider to be supported by the results of experiments in which $\alpha - $ particles were scattered by gold foil?

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$A$ beam of fast-moving alpha particles was directed towards a thin gold foil. The parts $A, B$,and $C$ of the incident beam and their corresponding transmitted or reflected parts $A^{\prime}, B^{\prime}$,and $C^{\prime}$ are shown in the adjoining diagram. The number of alpha particles in:

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What is the impact parameter?

$\sqrt{d_{1}}$ and $\sqrt{d_{2}}$ are the impact parameters corresponding to scattering angles $60^{\circ}$ and $90^{\circ}$ respectively,when an $\alpha$-particle is approaching a gold nucleus. For $d_{1} = x d_{2}$,the value of $x$ will be . . . . . .

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