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Question

An α particle with a kinetic energy of 1MeV is projected towards a stationary nucleus with a charge |75e|. Neglecting the motion of the nucleus, determine the distance of closest approach of the α particle.

A
2.14×104m
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B
2.14×108m
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C
2.14×1012m
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D
2.14×1014m
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Solution

The correct option is D 2.14×10−14mGiven, Kinetic energy =1μeV=106×1.6×10−19JoulesQ= charge of nucleus =75eq= charge of α−particle =2eWe have to find the distance of closest approach, we use the formula.Kinetic energy =14πε0Qqrr= distance of closest approachr=14πε0QqK.EOn substituting the values, =9×109×75×2×(1.6×10−19)21.6×10−19×106=2.14×10−14m

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