# Distance of closest approach

A proton and an $$\alpha$$-particle are projected with velocity $$v =\sqrt { \frac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }mL } }$$ each from infinity as shown.The perpendicular distance between their initial velocities is $$L$$.Let the distance of their closest approach be $$D$$.

Given $$x={ (\frac { 8D }{ L } -5) }^{ 2 }$$.Find $$x$$

Details and assumptions

1)mass of proton=$$m$$,charge=$$+e$$

2)mass of $$\alpha$$-particle=$$4m$$,charge=$$+2e$$

3)$${ \varepsilon }_{ 0 }$$=Vacuum Permittivity

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