# NEET Physics Atoms Questions Solved

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In a Geiger - Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when an $\mathrm{\alpha }$ particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction.

How will the distance of closest approach be affected when the kinetic energy of the $\mathrm{\alpha }$ is doubled?

$\frac{\left(\mathrm{ze}\right)\left(2\mathrm{e}\right)}{4{\mathrm{\pi \epsilon }}_{0}\left({\mathrm{r}}_{0}\right)}=\mathrm{k}\phantom{\rule{0ex}{0ex}}{\mathrm{r}}_{0}=\frac{2{\mathrm{Ze}}^{2}}{4{\mathrm{\pi \epsilon }}_{0}\mathrm{K}}$                                                                  (1) marks

${\mathrm{r}}_{0}=\frac{9×{10}^{9}×2×80×\left(1.6×{10}^{-19}{\right)}^{2}}{8×{10}^{6}×\left(1.6×{10}^{-19}\right)}\mathrm{m}$                              (1/2) marks

$=\frac{18×1.6×{10}^{-10}×80}{8×{10}^{6}}$                                                (1/2) marks

(1/2) marks

${\mathrm{r}}_{0}\propto \frac{1}{\mathrm{KE}}$

If KE becomes twice then ${\mathrm{r}}_{0}^{\text{'}}=\frac{{\mathrm{r}}_{0}}{2}$

i.e. distance of closest approach becomes half                   (1/2) marks

Difficulty Level:

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