An α-particle of kinetic energy \(K\) moves head-on towards a nucleus of atomic number 80. Its distance of closest approach is \(d\). Another α-particle of kinetic energy \(2K\) moves head-on towards a nucleus of atomic number 40. Its distance of closest approach is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(d/4\)
At the distance of closest approach, the α-particle stops for a moment. Its kinetic energy has all changed into electric potential energy. This gives \( d=2Ze^{2}/4\pi\varepsilon_0K \), so \( d \) is proportional to \( Z/K \).
\( \displaystyle d\propto \frac{Z}{K} \)
\( \begin{aligned} \frac{d^{\prime}}{d} &= \frac{40}{80}\times\frac{K}{2K} \\ &= \frac{1}{4} \end{aligned} \)
From the above, \( d^{\prime}=d/4 \). Hence, the distance of closest approach becomes one-fourth of \( d \).
NCERT (2026-27):
Example 12.2 – “Thus the distance of closest approach d is given by”
\( \displaystyle d=\frac{2Ze^{2}}{4\pi\varepsilon_0K} \)