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13.Nuclei
medium
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
$\frac{1}{m^2}$
D
$\frac{1}{{{v^2}}}$
Solution
Work done to stop the $\alpha$ particle is equal to $K.E.$
$\therefore q V=\frac{1}{2} m v^{2} \Rightarrow q \times \frac{K(Z e)}{r}=\frac{1}{2} m v^{2}$
$\Rightarrow r=\frac{2(2 e) K(Z e)}{m v^{2}}=\frac{4 K Z e^{2}}{m v^{2}}$
$\Rightarrow r \propto \frac{1}{v^{2}}$ and $r \propto \frac{1}{m}$
Standard 12
Physics