An atom ⁸₃X 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:

Explanation

Initially, the given atom has mass number 8. Electrons do not affect the nucleus, so only absorbed protons and neutrons will change the mass number.

In 10 seconds, the atom absorbs 10 protons and 9 neutrons. Hence, increase in mass number = 10 + 9 = 19.

Final mass number of the nucleus becomes:
8 + 19 = 27

The radius of a nucleus varies as the cube root of its mass number:
r ∝ A^1/3

Surface area of nucleus depends on square of radius:
Surface area ∝ r² ∝ A^2/3

Therefore, ratio of final to initial surface area is:
(27 / 8)^2/3

(27 / 8)^2/3 = (3 / 2)² = 9 / 4

Percentage increase in surface area:
= (9/4 − 1) × 100
= (5/4) × 100
= 250%

Hence, the correct answer is 250%.

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