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12.Atoms
medium
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
A
$v^2$
B
$\frac{1}{{Ze}}$
C
$\frac{1}{m}$
D
$\;\frac{1}{{{v^4}}}$
(AIPMT-2010) (AIEEE-2006)
Solution
For closest approach, kinetic energy is converted into potential energy
$\frac{1}{2}m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r_0}}}$ $ = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{(Ze)(2e)}}{{{r_0}}}$
${r_0} = \frac{{4Z{e^2}}}{{4\pi {\varepsilon _0}m{v^2}}} = $ $\frac{{Z{e^2}}}{{\pi {\varepsilon _0}{v^2}}}\left( {\frac{1}{m}} \right)$
$r_{0}$ is proportional to $\left(\frac{1}{m}\right)$
Standard 12
Physics
