Answer the following. (a) Define 'Mass defect' and 'Binding energy' of a nucleus. Describe 'Fission process' on the basis of binding energy per nucleon. (b) A deuteron contains a proton and a neutron and has a mass of 2.013553 u. Calculate the mass defect for it in u and its energy equivalence in MeV. (m_p = 1.007277 u, m_n = 1.008665 u, 1u = 931.5 MeV/c²)

PhysicsCBSE Class 12 · Board exam 2025, set 55/1/1 · 3 marks · Short answer

CBSE Class 12 Physics 2025: question 27

Set in the CBSE Class 12 board exam of 2025, paper 55/1/1, section C, question 27, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q27 · 3 marks · Short answerNuclei

Answer the following.

(a)
Define 'Mass defect' and 'Binding energy' of a nucleus. Describe 'Fission process' on the basis of binding energy per nucleon.
0 marks
(b)
A deuteron contains a proton and a neutron and has a mass of 2.013553 u. Calculate the mass defect for it in u and its energy equivalence in MeV. ( $m_{p}$ = 1.007277 u, $m_{n}$ = 1.008665 u, 1u = 931.5 $MeV/c^{2}$ )
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/1/1.

Hints

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

Hint 1

Compare the sum of the free proton and neutron masses with the deuteron's mass.

Hint 2

Multiply the mass defect in u by 931.5 MeV. For fission, think about where heavy and middle nuclei lie on the $E_{bn}$ vs A curve.

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.

(a) Mass defect: the difference between the total mass of the separate nucleons and the actual mass of the nucleus, $Δ M = [Z m_{p} + (A - Z) m_{n}] - M$ .
½ mark
Binding energy: the minimum energy needed to separate a nucleus completely into its constituent nucleons, $E_{b} = Δ M c^{2}$ .
½ mark
Fission: a heavy nucleus (A ≈ 240, $E_{bn}$ ≈ 7.6 MeV) splits into two medium nuclei (A ≈ 120, $E_{bn}$ ≈ 8.5 MeV). The binding energy per nucleon increases, so the nucleons become more tightly bound and the difference (≈ 200 MeV per fission) is released.
½ mark
(b) $Δ m = (m_{p} + m_{n}) - m_{d} = (1.007277 + 1.008665) - 2.013553$ u $= 2.015942 - 2.013553 = 0.002389$ u.
1 mark
Energy equivalent $= Δ m \times 931.5$ MeV/u $= 0.002389 \times 931.5 \approx 2.225$ MeV ( $\approx 2.22$ MeV).
½ mark

Answer: Δm = 0.002389 u; energy ≈ 2.225 MeV (≈ 2.22 MeV)

Where the marks are

Defines mass defect (½ mark)
Defines binding energy (½ mark)
Explains fission as an increase in binding energy per nucleon that releases energy (½ mark)
Calculates $Δ m = 0.002389$ u (1 mark)
Gets energy ≈ 2.22 MeV (½ mark)

Revise the chapter

Chapter: Nuclei (notes, formulas, where marks are lost and every past question)
Topic: Mass Defect and Binding Energy

Practise this chapterBoard-style questions from Nuclei, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2025: question 27

Hints

Step-by-step solution

Revise the chapter

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