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Question #43707

Difficulty: 8/10 | 🟢 Free

The Question

A hypothetical super-heavy nucleus of mass M=2hcGM = 2 \sqrt{\frac{hc}{G}} undergoes fission into two identical fragments. The mass defect Δm\Delta m of this process is given by Δm=35GM2Rc2\Delta m = \frac{3}{5} \frac{G M^2}{R c^2}, where the nucleus radius is R=h2πMcR = \frac{h}{2\pi M c}. Here hh is the Planck constant, cc is the speed of light, and GG is the universal gravitational constant.

A new system of units is proposed where the fundamental base units are:

  1. Action: A0=hA_0 = h
  2. Power: P0=c52GP_0 = \frac{c^5}{2G}
  3. Force: F0=c4GF_0 = \frac{c^4}{G}

What is the numerical value of the energy released in the fission process in this new system of units?

Options

  • A.

    9.6π29.6\pi \sqrt{2}

  • B.

    2.4π22.4\pi \sqrt{2}

  • C.

    4.8π24.8\pi \sqrt{2}

  • D.

    19.2π219.2\pi \sqrt{2}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Nuclear Fission Energy Balance|Derived Units via Fundamental Constants

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to maintain dimensional consistency when transitioning between standard algebraic expressions and custom unit systems.

The primary trap lies in the misapplication of the mass-energy equivalence principle (E=Δmc2E = \Delta m c^2) in non-SI units.

Students accustomed to natural units often erroneously assume the numerical value of cc is 1 in any system based on fundamental constants.

If they first convert the mass defect into the new mass unit (M0=22hc/GM_0 = 2\sqrt{2}\sqrt{hc/G}), they obtain a numerical mass defect of 2.4π22.4\pi\sqrt{2}.

Blindly setting E=ΔmE = \Delta m yields this incorrect value (Option B).

The rigorous approach requires either converting cc to the new units (Nc=2N_c = 2) or, more elegantly, deriving the new energy unit directly from Action and Power (E0=A0P0E_0 = \sqrt{A_0 P_0}), completely bypassing the mass conversion trap and the distractor Force unit.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterNuclei - Class 12
ConceptNuclear Fission Energy Balance
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree