Question

An $\alpha$ - particle accelerated through V volt is fired towards a nucleus.Its distance of closest approach is r.If a proton is accelerated through the same potential and fired towards the same nucleus,the distance of closest approach of proton will be:

• A. r
• B. 2r
• C. r/2
• D. r/4

Answer 1

Answer: (A)

r

Answer 2

K.E = P.E $\Rightarrow$ $q_1 V = \frac{1}{4 \pi \varepsilon_0} \frac{q_1q_2}{r}$
Independent of charge of the particle projected