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In alpha-particle scattering experiment,...

In alpha-particle scattering experiment, the perpendicular distance(b) of initial velocity vector of the the alpha-particle from the centre of nucleus is equal to (where E is initial kinetic energy of alpha-particle) (##AAK_MCP_38_NEET_PHY_E38_020_Q01##)

A

`(Ze^2sin^2(theta/2))/(4πepsilon_0E)`

B

`(Ze^2cot(theta/2))/(4πepsilon_0E)`

C

`(Ze^2cos(theta/2))/(4πepsilon_0E)`

D

`(Z^2e^2E)/(sin^4(theta/2)4πepsilon_0)`

Text Solution

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The correct Answer is:
B
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