Assertion (A) : The distance of closest approach to a gold nucleus of an $α$ -particle of K.E 5.5Mev is about 4.0 × 10^–14m.
Reason (R) : Rutherford by assuming that the coulomb repulsive force was only responsible for scattering.

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(A) is true but (R) is false

Both (A) and (R) are true and (R) is the correct explanation of (A)

Both (A) and (R) are false

Both (A) and (R) are true and (R) is not the correct explanation of (A)

answer is B.

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Detailed Solution

for distance of closest approach

$\frac{Ze 2 e}{4 {πε}_{0} r} = \frac{1}{2} {mv}^{2} where r is the distance of closest approach$

for gold nucleus

$\frac{79 \times 1.6 \times 10^{- 19} \times 2 \times 1.6 \times 10^{- 19}}{r} \times 9 \times 10^{9} =$

$5.5 \times 1.6 \times 10^{- 13}$

$solving for r we get r = 4.0 \times 10^{- 14} m$

scattering according to Rutherford is due to the repulsion of $α$ particle with the nucleus.

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