Distance of closest approach

A proton and an α\alpha -particle are projected with velocity v=e24πε0mL v =\sqrt { \frac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }mL } } each from infinity as shown.The perpendicular distance between their initial velocities is LL.Let the distance of their closest approach be DD.

Given x=(8DL5)2x={ (\frac { 8D }{ L } -5) }^{ 2 }.Find xx

Details and assumptions

1)mass of proton=mm,charge=+e+e

2)mass of α\alpha -particle=4m4m,charge=+2e+2e

3)ε0{ \varepsilon }_{ 0 }=Vacuum Permittivity

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