JEE Mains Physics · Nuclei

Q-value, Fission and Fusion

The energy a nuclear reaction releases, its Q-value, is the gain in total binding energy, or equally the mass lost times c²; one reaction's Q times the number of nuclei gives the energy of a whole sample.

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Why this matters

Twenty-three PYQs, ten of them numerical, and six from 2026. Eight find the energy released from binding energies per nucleon, in a fission or a fusion. Ten find it from masses: alpha decay, a fission, a fusion chain, and how Q is shared between the products. Five scale one reaction up to grams of fuel or a star's power output.

Concept 1 of 3: Q from binding energies per nucleon

Binding energy is energy the nucleus has already given away. If the products are more tightly bound than what you started with, the extra binding energy comes out as released energy. So multiply each BE per nucleon by its mass number to get totals, and subtract: products minus reactants.

Definition

Total binding energy of a nucleus = $A \times (BE/A)$ .
$Q = \sum BE_{\text{products}} - \sum BE_{\text{reactants}}$ .
$Q > 0$ : energy released. $Q < 0$ : energy must be supplied.
A free proton or neutron has zero binding energy.
Fission: a heavy nucleus (about 7.6 MeV per nucleon) splits into two middle ones (about 8.5 MeV per nucleon), so roughly 1 MeV per nucleon, about 200 MeV per fission, is released.
Fusion: light nuclei join; the jump in BE per nucleon is large, so the energy per nucleon is larger than in fission.

Q-value from binding energy

Q = \sum_{\text{products}} A\,b - \sum_{\text{reactants}} A\,b

$b$ binding energy per nucleon of that nucleus

Worked example

A nucleus with A = 200 and binding energy 7.8 MeV per nucleon splits into two equal nuclei with binding energy 8.7 MeV per nucleon. Find the energy released.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q15Moderate

Example 1 · Nuclei · Q-value, Fission and Fusion

Nucleus

A

is having mass number 220 and its binding energy per nucleon is

5.6MeV

. It splits in two fragments '

B

' and '

C

' of mass numbers 105 and 115. The binding energy of nucleons in '

B

' and '

C

' is

6.4MeV

per nucleon. The energy

Q

released per fission will be:

Multiply by A before subtracting

BE per nucleon values cannot be subtracted directly: 8.4 − 7.6 = 0.8 MeV is per nucleon, not the answer. Multiply each by its own A first, then subtract totals.

Products minus reactants, for binding energy

With binding energies, Q = after − before. With masses it is the other way round, before − after. Mixing the two flips the sign.

Count every product nucleus

Two deuterons make one helium: the reactants hold 2 × 2 = 4 nucleons. Forgetting the second deuteron halves the reactant side.

Drill 7 more on q from binding energies per nucleon

Concept 2 of 3: Q from masses, and how it is shared

Mass that disappears in a reaction reappears as kinetic energy. Add up the masses before, subtract the masses after, and multiply by 931.5 MeV per u. In a decay at rest, momentum must stay zero, so the light alpha flies off fast and the heavy daughter recoils slowly: the alpha takes almost all of Q, but not quite all.

Definition

$Q = \left(\sum m_{\text{reactants}} - \sum m_{\text{products}}\right)c^{2}$ , with $1\ \text{u}\,c^{2} = 931.5$ MeV.
Atomic masses can be used when the electron count is the same on both sides.
A chain of reactions: add the steps; an intermediate made in one step and used in the next cancels.
Q goes into kinetic energy: $K_{\text{products}} = K_{\text{projectile}} + Q$ . The products' energy cannot be negative, so $K_p + Q > 0$ .
Alpha decay of a nucleus at rest: $K_\alpha = Q\dfrac{A - 4}{A}$ , daughter $K_d = Q\dfrac{4}{A}$ .
Splitting at rest into equal pieces: momenta cancel, and the total kinetic energy equals $\Delta m\,c^{2}$ .

Q-value from masses; alpha's share

Q = \left(\sum m_i - \sum m_f\right)c^{2}, \qquad K_\alpha = Q\,\frac{A - 4}{A}

Worked example

^{226}\text{Ra} \rightarrow {}^{222}\text{Rn} + {}^{4}\text{He}

. Atomic masses: Ra 226.025410 u, Rn 222.017578 u, He 4.002603 u; 1 u = 931.5 MeV/

c^{2}

. Find Q and the alpha's kinetic energy.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q20Moderate

Example 2 · Nuclei · Q-value, Fission and Fusion

\ _{92}^{238}\text{ }A \rightarrow\ _{90}^{234}\text{ }B +\ _{2}^{4}D + Q

In the given nuclear reaction, the approximate amount of energy released will be : [Given, mass of

\ _{92}^{238}\text{ }A = 238.05079 \times 931.5MeV/c^{2}

, mass of

\ _{90}^{234}\text{ }B = 234.04363 \times 931.5MeV/c^{2}

, mass of

\ _{2}^{4}D = 4.00260 \times 931.5MeV/c^{2}

]

The alpha does not get all of Q

Momentum is shared equally and oppositely, so kinetic energy splits inversely with mass. The alpha gets

Q(A - 4)/A

. Giving it all of Q ignores the recoil.

Before minus after, for masses

Mass lost is released energy, so Q = (reactant masses) − (product masses). A negative answer means the reaction needs energy.

Keep the electrons balanced

Atomic masses carry electrons. In alpha decay the parent atom's electrons equal the daughter's plus helium's, so atomic masses work. In beta-plus decay they do not balance; take care there.

Drill 9 more on q from masses, and how it is shared

Concept 3 of 3: Energy from a sample, and power

One fission gives about 200 MeV, a tiny amount. A gram of uranium holds about 2.6 × 10²¹ nuclei, so the total is huge. Count the nuclei with moles and Avogadro's number, multiply by the energy per reaction, and convert MeV to joules if asked. For a reactor or a star, divide the power by the energy per reaction to get reactions per second.

Definition

Number of nuclei $N = \dfrac{m}{M}N_A$ , with m and the molar mass M in the same unit.
Total energy $E = N \times Q$ .
$1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J}$ .
Power: reactions per second $= \dfrac{P}{Q}$ , with Q in joules. Mass used per second = (reactions per second) × (mass per reaction).
When several nuclei make one reaction (three helium into one carbon), divide the count of nuclei by that number first.

Energy from a sample

E = \frac{m}{M}N_A\,Q, \qquad \text{rate} = \frac{P}{Q}

Worked example

Each fission of

^{235}\text{U}

releases 200 MeV. Find the energy, in joules, if every nucleus in 1 kg of

^{235}\text{U}

undergoes fission. (

N_A = 6.022 \times 10^{23}

, 1 MeV =

1.6 \times 10^{-13}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q21Moderate

Example 3 · Nuclei · Q-value, Fission and Fusion

The average energy released per fission for the nucleus of

\ _{92}^{235}U

is 190 MeV. When all the atoms of 47 g pure

\ _{92}^{235}U

undergo fission process, the energy released is

\alpha \times 10^{23}MeV

. The value of

\alpha

\_\_\_\_

. (Avogadro Number

= 6 \times 10^{23}

per mole)

Grams over grams per mole

Moles = mass ÷ molar mass with both in grams. Using kilograms for one and grams for the other is off by a thousand.

MeV is not joules

Power is in watts, joules per second. Turn Q into joules (× 1.6 × 10⁻¹³) before dividing a power by it.

Several nuclei per reaction

If three helium nuclei make one carbon, the number of reactions is a third of the number of helium nuclei.

Drill 4 more on energy from a sample, and power

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Q from binding energies per nucleon
Q-value from binding energy
$Q = \sum_{\text{products}} A\,b - \sum_{\text{reactants}} A\,b$
Q from masses, and how it is shared
Q-value from masses; alpha's share
$Q = \left(\sum m_i - \sum m_f\right)c^{2}, \qquad K_\alpha = Q\,\frac{A - 4}{A}$
Energy from a sample, and power
Energy from a sample
$E = \frac{m}{M}N_A\,Q, \qquad \text{rate} = \frac{P}{Q}$

Watch out for (9)

Multiply by A before subtracting
→ Q from binding energies per nucleon
Products minus reactants, for binding energy
→ Q from binding energies per nucleon
Count every product nucleus
→ Q from binding energies per nucleon
The alpha does not get all of Q
→ Q from masses, and how it is shared
Before minus after, for masses
→ Q from masses, and how it is shared
Keep the electrons balanced
→ Q from masses, and how it is shared
Grams over grams per mole
→ Energy from a sample, and power
MeV is not joules
→ Energy from a sample, and power
Several nuclei per reaction
→ Energy from a sample, and power

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