JEE Main · 2006 · 30 Apr 2006 · Offline · Electrostatics

An alpha nucleus of energy

\frac{1}{2} m v^{2}

bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

(A) $\frac{1}{Ze}$
(B) $v^{2}$
(C) $\frac{1}{m}$ ✓ Correct
(D) $\frac{1}{v ^{4}}$

Answer: (C)

Solution

At the distance of closest approach the alpha particle has momentarily stopped, so all its kinetic energy has turned into electrostatic potential energy:

\frac{1}{2} m v^{2} = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{d}

. Solving,

d = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{\frac{1}{2} m v ^{2}}

, so for a fixed speed the closest-approach distance is proportional to

\frac{1}{m}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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