Question

An atom \( ^8_3X \) is bombarded by a 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 nucleus is recorded by:

• A. \(150%\)
• B. \(250%\)
• C. \(900%\)
• D. \(225%\)

Hint:

Electrons do not contribute to nuclear size; only protons and neutrons affect the mass number.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The size of a nucleus depends solely on its mass number \(A\). Electrons do not reside in the nucleus and do not contribute to the nuclear surface area.
Step 2: Key Formula or Approach:
Nuclear radius: \(R = R_0 A^{1/3}\).
Nuclear surface area: \(S = 4\pi R^2 \propto A^{2/3}\).
Step 3: Detailed Explanation:
Initial atom is \({}^8_3 X\), so the initial mass number \(A_i = 8\).
Particles absorbed that add to the nucleus: 10 protons and 9 neutrons.
Final mass number \(A_f = A_i + 10 + 9 = 8 + 19 = 27\).
Initial surface area \(S_i \propto (A_i)^{2/3} = (8)^{2/3} = (2^3)^{2/3} = 2^2 = 4\).
Final surface area \(S_f \propto (A_f)^{2/3} = (27)^{2/3} = (3^3)^{2/3} = 3^2 = 9\).
Percentage growth recorded (calculated as final area relative to initial area based on options):
\[ \text{Percentage} = \frac{S_f}{S_i} \times 100% = \frac{9}{4} \times 100% = 2.25 \times 100% = 225% \]
Step 4: Final Answer:
The percentage growth in the surface area recorded is \(225%\).