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NEET - 2010

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

(a) $\frac{1}{Ze}$                              (b) ${v}^{2}$

(c) $\frac{1}{m}$                               (d) $\frac{1}{{v}^{4}}$

$\alpha ‐$particle of mass m possesses initial velocity 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,

$\frac{1}{2}m{v}^{2}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{\left(2e\right)\left(Ze\right)}{{r}_{0}}$

${r}_{0}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{2Z{e}^{2}}{\frac{1}{2}m{v}^{2}}$

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