Find the closest distance of approach.

1.4 A

^{0}

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2.8 A

^{0}

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2.8 p m

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2.8 f m

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Solution

The correct option is C $2.8 A$ $^{0}$
the kinetic energy of particle is converted into interaction(repulsive) energy
${K . E .}_{i n i t i a l} + {P . E}_{i n i t i a l} = {P . E}_{f i n a l}$

$\frac{1}{2} m v^{2} + F . x_{i} = F . x_{f}$

$\frac{1}{2} \times (1.67 \times 10^{- 27}) \times {(3 \times 10^{5})}^{2} + \frac{2.12 \times 10^{- 26}}{5^{2}} \times 5 = \frac{2.12 \times 10^{- 26}}{x^{2}} \times x$

by solving this we get

$x = \frac{2.12 \times 2 \times 10^{- 9}}{1.67 \times 9} = 2.8 \times 10^{- 10} = 2.8 ˙ A$

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Find the closest distance of approach.

The correct option is C 2.8A0the kinetic energy of particle is converted into interaction(repulsive) energyK.E.initial+P.Einitial=P.Efinal12mv2+F.xi=F.xf12×(1.67×10−27)×(3×105)2+2.12×10−2652×5=2.12×10−26x2×xby solving this we getx=2.12×2×10−91.67×9=2.8×10−10=2.8˙A