jee-main 2006 Q46

jee-main · India Not Maths

An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to
(1) $\frac{1}{Ze}$
(2) $v^2$
(3) $\frac{1}{m}$
(4) $\frac{1}{v^4}$

An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to\\
(1) $\frac{1}{Ze}$\\
(2) $v^2$\\
(3) $\frac{1}{m}$\\
(4) $\frac{1}{v^4}$

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