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Question

# The closest distance of approach of an α− particle travelling with a velocity v towards Al(Z=13) nucleus is d. The closest distance of approach of an α−particle travelling with velocity 4v towards Fe(Z=26) nucleus is-

A
d16
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B
d8
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C
d4
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D
d2
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Solution

## The correct option is B d8At the distance of closest approach, K.Eα=Electrostatic P.Eα 12mv2=14πε0q1q2r=14πε02Ze2r [∵ q1=Ze ; q2=2e] ∴ r∝Zv2 ⇒ r2r1=Z2Z1×[v1v2]2 Substituting the given data gives, ⇒ r2d=2613×[v4v]2=2×116 [∵ r1=d] ⇒ r2=d8 <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (B) is the correct answer.

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