7.9 MeV α-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) ______ m. [4πϵ01=9×109 Nm²/C² and electron charge =1.6×10−19 C]
(A)5.76×10−14✓ Correct
(B)1.44×10−13
(C)2.88×10−14
(D)1.69×10−12
Answer: (A)
Solution
At the distance of closest approach the α-particle is momentarily at rest and its kinetic energy has wholly become electrostatic potential energy: K=4πε01r(2e)(Ze). Solving, r=7.9×106×1.6×10−19(9×109)(2)(79)(1.6×10−19)2=2.88×10−14m. This separation estimates the nuclear radius, so the estimated diameter is 2r=5.76×10−14m.
JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option
