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PhysicsCBSE Class 12 · Sample paper, session 2025-26 · 3 marks · Short answer

CBSE Class 12 Physics sample paper 2025-26: question 24

From the official CBSE sample question paper for 2025-26, section C, question 24, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q24 · 3 marks · Short answerNuclei

A fast-moving neutron collides with the nucleus of Plutonium (Pu), thereby producing Xenon (Xe) and Zirconium (Zr) along with neutrons.
(I) Write the nuclear fission reaction.
(II) Find the energy released in the above nuclear reaction.
Given atomic masses:
m() = 239.052157u,
m() = 102.926597u,
m() = 133.905040u &
m() = 1.00866u.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

Open one hint at a time and try again before the next.

Hint 1

Balance mass numbers and atomic numbers to find how many neutrons are released.

Hint 2

Mass defect = total initial mass − total final mass.

Hint 3

1 u is equivalent to 931.5 MeV.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (I) (mass number ; charge ).

    1 mark
  2. (II) Mass defect:

  3. 1 mark
  4. Energy released

    1 mark

Answer: (I) Pu-239 + n → Xe-134 + Zr-103 + 3n; (II) Δm = 0.2032 u, Q ≈ 189.3 MeV

Where the marks are

  • Correct balanced fission equation with 3 neutrons (1 mark)
  • Mass defect 0.2032 u (1 mark)
  • Energy released ≈ 189.3 MeV (1 mark)

Revise the chapter

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.