KVPY SA Exam 2018 Question Paper

Question : 69 of 80

Marks: +1, -0

A proton of mass m and charge e is projected from a very larger distance towards an α particle with velocity v. Initially, α particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be

$e^2 /{4πε_0mv ^2}$
${5e^2} /{4πε_0mv^ 2}$
${ 2e^2 }/{4πε∈_0mv ^2}$
${ 4e^2 }/{4πε_0 mv ^2}$

Validate

Solution:

1/2({m.4m}/{m+4m})v^2=1/{4πε_0}\ {2e^2}/r

r= {5e^2} /{4πε_0mv^ 2}

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