Question

An elementary particle of mass m' and
charge +e initially at a very large distance is
projected with velocity 'y' at a much more
massive particle of charge +Ze at rest. The
closest possible distance of approach of the
incident particle is
Ze?
41 E, mV
Ze²
2n Eo
mV
Ze?
TI E, mV
Ze
211 €, mV​

Answer 1

Given:

Elementary particle of mass m and charge e is projected with velocity from a large distance towards a massive particle of charge Ze at rest.

To find:

Closest distance of approach of mass m.

Calculation:

The closest distance will be achieved when all the Kinetic energy of the incident particle will be converted into the Potential Energy.

Let distance of closest approach be d

$\sf{kinetic \: energy = potential \: energy}$

$\sf{ = > \dfrac{1}{2} m {v}^{2} = \dfrac{k(e)(Ze)}{d} }$

$\sf{ = > d = \dfrac{2k(e)(Ze)}{m {v}^{2} } }$

Putting value of Coulomb's Constant:

$\sf{ = > d = \dfrac{2(e)(Ze)}{4\pi\epsilon_{0}m {v}^{2} } }$

$\sf{ = > d = \dfrac{(e)(Ze)}{2\pi\epsilon_{0}m {v}^{2} } }$

$\sf{ = > d = \dfrac{Z {e}^{2} }{2\pi\epsilon_{0}m {v}^{2} } }$

So final answer is :

$\boxed{ \sf{ d = \dfrac{Z {e}^{2} }{2\pi\epsilon_{0}m {v}^{2} } }}$