CBSE 2022 · Region 1 · Set 1 · Q4 · 3 marks
Define the term - Distance of closest approach. How will it be affected, for an $\displaystyle \alpha$ - particle, if kinetic energy of the particle is doubled ?
Marking-scheme solution
The minimum distance up to which an alpha particle travel along the central line of the nucleus before it rebounds is called distance of closest approach.
Alternatively
An alpha particle travelling directly towards the centre of a nucleus slows down as it approaches the nucleus due to repulsive force. At a distance $\displaystyle \mathrm{r}_{0}$ from the nucleus, the $\displaystyle \alpha$ - particle stops and its total kinetic energy converts into electrostatic potential energy. This distance $\displaystyle \mathrm{r}_{0}$ is called distance of closest approach.r_{0}=\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} K . E .}i.e., $\displaystyle r_{0} \propto \frac{1}{\text { K.E. }}$
as K is doubled, $\displaystyle \mathrm{r}_{0}$ is halved.
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.