JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Modern Physics

An atom

_{3}^{8} X

is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleons is recorded by:

(A) $250%$
(B) $150%$
(C) $225%$ ✓ Correct
(D) $900%$

Answer: (C)

Solution

The nucleus of

_{3}^{8} X

has

A = 8

nucleons. Electrons are not nucleons, so only the 10 absorbed protons and 9 absorbed neutrons increase the nucleon count:

A_{f} = 8 + 10 + 9 = 27

. Since

R = R_{0} A^{1/3}

, the surface area

S \propto R^{2} \propto A^{2/3}

. Thus

\frac{S _{f}}{S _{i}} = (\frac{27}{8})^{2/3} = (\frac{3}{2})^{2} = \frac{9}{4} = 2.25

; the surface area becomes

225%

of its original value.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option

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