Nuclear Forces and Nuclear Energy

PhysicsNucleiFor JEE aspirants

Nuclear forces and nuclear energy answer two questions: what holds protons and neutrons together, and how that binding releases energy. The nuclear force is the strongest force in nature but acts only over about $2 fm$ . Splitting a heavy nucleus (fission) or joining light nuclei (fusion) moves nucleons to more tightly bound nuclei, releasing about $200 MeV$ per fission and $17.6 MeV$ per D-T fusion. Nuclear forces and nuclear energy appear in JEE Main and NEET every year.

On this page1Nuclear force2Potential energy curve3Saturation4Nuclear fission5Chain reaction6Nuclear reactor7Nuclear fusion8Energy in stars9Controlled fusion10Fission vs fusion

Key Formulas - Quick Reference

★ Must learnNuclear force: strongest ( $\approx 100$ times Coulomb at $1 fm$ ), short range ( $\approx 2$ to $3 fm$ ), charge independent ( $F_{pp} \approx F_{nn} \approx F_{n p}$ ), saturating, attractive for $r > 0.8 fm$ and repulsive below
★ Must learnFission: $_{92}^{235} U +_{0}^{1} n \to_{92}^{236} U^{*}$ , which splits as $\to_{56}^{141} Ba +_{36}^{92} Kr + 3_{0}^{1} n + Q$ ; $Q \approx 200 MeV$
$Q = [(m_{U} + m_{n}) - (m_{B a} + m_{K r} + 3 m_{n})] \times 931.5 MeV$ (atomic masses; electrons cancel)
Chain reaction: multiplication factor $k$ ; $k < 1$ subcritical, $k = 1$ critical (reactor), $k > 1$ supercritical (bomb); $N = N_{0} k^{n}$
★ Must learnFusion: $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He + n + 17.6 MeV$ ; $^{2} H +^{2} H \to^{3} He + n + 3.27 MeV$ or $^{3} H + p + 4.03 MeV$
Coulomb barrier $U = \frac{k Z _{1} Z _{2} e ^{2}}{R _{1} + R _{2}}$ , with $k e^{2} = 1.44 MeV fm$ ; thermal energy $\frac{3}{2} k_{B} T$
★ Must learnSun (p-p chain): $4_{1}^{1} H \to_{2}^{4} He + 2 e^{+} + 2 ν + 26.7 MeV$
Two-body break-up at rest: $K_{1} : K_{2} = m_{2} : m_{1}$ , $K_{1} = \frac{m _{2}}{m _{1} + m _{2}} Q$

1. The Nuclear Force

A nucleus packs up to about $90$ protons into a sphere a few femtometres across. At $1 fm$ two protons repel with a Coulomb force of $\frac{k e ^{2}}{r ^{2}} \approx 230 N$ , huge for particles of mass $1 0^{- 27} kg$ . Gravity between them is about $1 0^{36}$ times weaker still. So there must be a much stronger attractive force between nucleons: the nuclear force (the strong interaction).

Properties of the nuclear force

Strongest force in nature. Within nuclear distances it is about $100$ times the Coulomb force: $F_{n} \approx 100 F_{e}$ .
Short range. It is very strong up to about $2$ to $3 fm$ and falls to nearly zero beyond a few fermi, so it acts only inside the nucleus.
Mainly attractive, repulsive at very short range. Attraction holds nucleons together; below about $0.8 fm$ it becomes strongly repulsive, which keeps nucleons from collapsing into each other.
Charge independent. It is almost the same between $n$ - $n$ , $n$ - $p$ and $p$ - $p$ pairs; it does not depend on electric charge.
Saturating. A nucleon interacts only with its nearest neighbours.
Spin dependent and non-central. It is stronger when the spins of the two nucleons are parallel, and it depends on their orientation, not only on the distance between them.

Force	Relative strength	Range	Acts between
Strong nuclear	$1$	$\approx 1 0^{- 15} m$	nucleons (quarks)
Electromagnetic	$1 0^{- 2}$	infinite	charged particles
Weak nuclear	$1 0^{- 13}$	$< 1 0^{- 16} m$	leptons and quarks ( $β$ -decay)
Gravitational	$1 0^{- 39}$	infinite	all masses

1.1 Potential energy of two nucleons

The potential energy of a pair of nucleons as a function of their separation shows all of this in one curve.

Figure 1: Potential energy of two nucleons (top, schematic shape; well depth and

r_{0}

are typical values) and the force

F = - \frac{d U}{d r}

computed from it (bottom).

F = 0

at the minimum

r_{0} \approx 0.8 fm

; it is repulsive inside

r_{0}

, attractive outside, strongest near 1.0 fm, and negligible beyond about

2.5 fm

The minimum of the potential energy is near $r_{0} \approx 0.8 fm$ . For $r > r_{0}$ the energy rises as the nucleons separate, so the force pulls them together; for $r < r_{0}$ it rises steeply as they approach, so the force pushes them apart. Nucleons in a nucleus sit roughly $r_{0}$ apart. The lower panel is the force $F = - \frac{d U}{d r}$ read from the slope of the upper one: it is zero at the minimum, positive (repulsive) where $U$ falls with $r$ , and most strongly attractive where $U$ rises most steeply, near $1 fm$ . The Coulomb energy of two protons (dashed) is tiny on this scale, which is why the nuclear force wins inside the nucleus.

1.2 Saturation and what it explains

Figure 2: Saturation. The nuclear force reaches only the touching neighbours (about

2 fm

), so each interior nucleon has a fixed number of bonds however big the nucleus:

B E \propto A

. Surface nucleons have fewer bonds, which lowers

\frac{B E}{A}

of small nuclei. Coulomb repulsion acts between all

\frac{Z ( Z - 1 )}{2}

proton pairs, which is why heavy nuclei need extra neutrons.

$B E \propto A$ and $\frac{B E}{A} \approx 8 MeV$ , nearly constant for medium nuclei (see Mass-Energy and Nuclear Binding Energy).
Constant density and $R \propto A^{1/3}$ : nucleons pack like molecules in a liquid drop.
Why heavy nuclei need extra neutrons: the Coulomb energy grows as $Z^{2}$ (every proton pair repels) while nuclear attraction grows only as $A$ . Extra neutrons add attraction without adding repulsion, so stable heavy nuclei have $N > Z$ (up to $\frac{N}{Z} \approx 1.5$ for lead). Beyond $Z = 83$ no nucleus is completely stable.

Why is the nuclear force so short-ranged? In Yukawa's theory (1935) nucleons attract by exchanging particles called pions (mass $\approx 140 MeV / c^{2}$ ). A virtual pion can exist only for a time $Δ t \approx \frac{ℏ}{m c ^{2}}$ , so it travels at most $c Δ t \approx \frac{ℏ}{m c} \approx 1.4 fm$ . That sets the range of the force. It is an exchange force, unlike gravity or electrostatics.

Quick Recall: tap to check

Name three properties of the nuclear force.

Any three: strongest force, short range (a few fm), charge independent, saturating, repulsive below about 0.8 fm, spin dependent, non-central.

Why can the nuclear force not be gravitational or electrical?

Gravity is about

1 0^{36}

times too weak, and the electrical force between protons is repulsive and does not act on neutrons.

Why do heavy stable nuclei have more neutrons than protons?

Coulomb repulsion grows as

Z^{2}

while nuclear attraction saturates; extra neutrons add attraction without repulsion.

Key idea

The nuclear force is strong, short-range and saturating: each nucleon grips only its neighbours, while every proton pushes on every other proton.

2. Nuclear Energy: Why Fission and Fusion Release Energy

The binding energy per nucleon curve peaks near $A = 56$ . Any process that moves nucleons from nuclei of low $\frac{B E}{A}$ into nuclei of higher $\frac{B E}{A}$ releases the difference as energy. There are two such routes:

Fission

A heavy nucleus ( $A > 200$ , $\frac{B E}{A} \approx 7.6 MeV$ ) splits into two middle nuclei ( $\approx 8.5 MeV$ ). Gain $\approx 0.9 MeV$ per nucleon, $\approx 200 MeV$ per event.

Fusion

Light nuclei ( $A < 20$ ) join into a heavier one far up the steep left side of the curve. D-T fusion gains $3.5 MeV$ per nucleon, $17.6 MeV$ per event.

Nuclear energies are about a million times larger than chemical ones: burning one carbon atom gives $4 eV$ , one fission gives $200 MeV$ .

3. Nuclear Fission

In 1938 Hahn and Strassmann found barium among the products of uranium bombarded with neutrons; Meitner and Frisch explained it as the splitting of the nucleus into two parts of comparable mass, and called it fission.

Nuclear fission is the splitting of a heavy nucleus ( $A > 200$ ) into two middle-mass fragments, with the release of $2$ or $3$ neutrons and about $200 MeV$ of energy. The most important example is uranium-235 hit by a slow (thermal) neutron:

_{92}^{235} U +_{0}^{1} n \to_{92}^{236} U^{*} \to_{56}^{141} Ba +_{36}^{92} Kr + 3_{0}^{1} n + Q

Figure 3: Fission as a liquid drop. The absorbed neutron's binding energy (

6.5 MeV

) excites

^{236} U^{*}

; it deforms, and once a neck forms the Coulomb repulsion of the two halves drives them apart. For this channel

Q = 173 MeV

from atomic masses.

Slow neutrons work best. Thermal neutrons (kinetic energy about $0.025 eV$ at room temperature) spend longer near the nucleus and are captured far more readily by $^{235} U$ . $^{238} U$ (99.3% of natural uranium) needs fast neutrons above about $1 MeV$ , and even then mostly captures them without splitting. Fission is also possible for $^{233} U$ and $^{239} Pu$ .
Many possible pairs. Other channels include $_{56}^{144} Ba +_{36}^{89} Kr + 3 n$ and $_{54}^{140} Xe +_{38}^{94} Sr + 2 n$ . On average about $2.5$ neutrons are released per fission, with energies near $2 MeV$ (fast neutrons).
Fragments are neutron-rich (they keep roughly the $\frac{N}{Z}$ of uranium), so they are radioactive and reach stability by a series of $β^{-}$ decays: radioactive waste.

3.1 Energy released in fission

Figure 4: Fission moves nucleons up the curve. Total binding energy before:

235 \times 7.59 = 1784 MeV

; after:

141 \times 8.33 + 92 \times 8.51 = 1957 MeV

(free neutrons have no binding energy), so

Q = 173 MeV

. The vertical axis starts at

7.0 MeV

With atomic masses $m (^{235} U) = 235.043928 u$ , $m (^{141} Ba) = 140.914404 u$ , $m (^{92} Kr) = 91.926173 u$ and $m_{n} = 1.008665 u$ , the mass lost in the reaction above is $0.1860 u$ , so $Q = 173 MeV$ . Averaged over all channels, including the later decays of the fragments, about $200 MeV$ is released per fission, which is about $0.2 u$ of mass.

Figure 5: One

^{235} U

fission. (a) The split is usually asymmetric, one fragment near

A \approx 95

and the other near

A \approx 139

(smooth approximation to measured yields; the groups add to

200%

because each fission gives two fragments). (b) Typical energy budget (rounded): over

80%

appears at once as kinetic energy of the fragments, which becomes heat in the fuel; the antineutrino share is lost.

Exam Trick

Fuel from power: number of fissions per second = P ÷ energy per fission. With $200 MeV = 3.2 \times 1 0^{- 11} J$ , a $1 W$ output needs $3.1 \times 1 0^{10}$ fissions per second. Complete fission of $1 kg$ of $^{235} U$ gives about $8.2 \times 1 0^{13} J$ , the energy of about $2500$ tonnes of coal.

Quick Recall: tap to check

Why are slow neutrons used for fission of

^{235} U

They are captured far more readily (they stay near the nucleus longer), so the probability of fission is much higher.

About how much energy is released per fission, and in what form mostly?

About

200 MeV

, over

80%

as kinetic energy of the fragments.

Why are fission fragments radioactive?

They are neutron-rich for their mass, so they undergo

β^{-}

decay.

4. Chain Reaction and the Nuclear Reactor

Each fission releases $2$ or $3$ neutrons. If at least one of them causes another fission, the process sustains itself: a chain reaction (first achieved by Enrico Fermi in 1942).

Figure 6: A chain reaction. The multiplication factor

k

is the average number of neutrons from one fission that cause another fission. Top: 3 neutrons per fission, 1 lost, so

k = 2

and the fissions double every generation. Bottom: after

n

generations there are

N_{0} k^{n}

fissions: dying out for

k < 1

, steady for

k = 1

, runaway for

k > 1

The multiplication factor $k = \frac{number of neutrons in one generation}{number in the previous generation}$ .

$k < 1$ : subcritical, the reaction dies out.
$k = 1$ : critical, steady rate, the condition in a nuclear reactor.
$k > 1$ : supercritical, the rate grows as $k^{n}$ ; if uncontrolled, an explosion (atom bomb).

Neutrons are lost by escaping through the surface and by capture without fission (for example in $^{238} U$ ). Loss through the surface falls as the size grows (surface $\propto R^{2}$ , production $\propto R^{3}$ ), so a chain reaction needs a minimum mass of fuel, the critical mass (about $50 kg$ for a bare sphere of pure $^{235} U$ ).

4.1 Parts of a nuclear reactor

Figure 7: A pressurised-water power reactor (schematic). The moderator slows neutrons, the control rods absorb them to keep

k = 1

, the primary coolant (kept liquid by high pressure) carries the fission heat to the steam generator, and the separate secondary loop turns a turbine and generator. The two loops never mix, so the radioactive water stays inside the shield.

Part	Material	Purpose
Fuel	Uranium enriched to about $3%$ $^{235} U$ (or natural U, $^{239} Pu$ )	Undergoes fission, source of energy
Moderator	Ordinary water, heavy water $(D_{2} O)$ , graphite	Slows fast neutrons ( $2 MeV$ ) to thermal energies by elastic collisions with light nuclei
Control rods	Cadmium or boron	Absorb neutrons; pushed in or pulled out to keep $k = 1$
Coolant	Water, heavy water, liquid sodium, $C O_{2}$	Carries the heat from the core to the steam generator
Shielding	Thick concrete and steel	Stops neutrons and gamma rays from reaching people
Reflector	Graphite or beryllium around the core	Returns escaping neutrons into the core

A good moderator has light nuclei (a neutron loses the most energy in a head-on collision with a nucleus of equal mass) that do not absorb neutrons. This is why ordinary hydrogen is efficient but absorbs some neutrons, while heavy water absorbs very few and allows natural uranium as fuel, as in India's pressurised heavy-water reactors. Fast breeder reactors use fast neutrons to turn $^{238} U$ into fissile $^{239} Pu$ , producing more fuel than they burn.

Figure 8: The moderator. In a head-on elastic collision a neutron loses the fraction

\frac{4 A}{( A + 1 ) ^{2}}

of its kinetic energy: all of it to a proton,

\frac{8}{9}

to a deuteron, only 28% to carbon. Averaged over all angles, slowing a

2 MeV

neutron to

0.025 eV

takes about 18 collisions in hydrogen, 25 in deuterium and 115 in graphite.

JEE Advanced

Energy lost in one head-on elastic collision. A neutron (mass $m$ , speed $v$ ) hits a nucleus of mass $A m$ at rest. Momentum and kinetic energy conservation give the neutron's final velocity

v^{'} = \frac{1 - A}{1 + A} v \Rightarrow \frac{K ^{'}}{K} = (\frac{A - 1}{A + 1})^{2}, \frac{Δ K}{K} = \frac{4 A}{( A + 1 ) ^{2}}

Hydrogen ( $A = 1$ ) can take all the energy in one collision, deuterium $\frac{8}{9}$ , carbon $\frac{48}{169} = 28%$ . Real collisions are mostly glancing, so the average number needed to go from $E_{0}$ to $E$ is $n = \frac{ln ( E _{0} / E )}{ξ}$ , where $ξ$ is the mean loss of $ln E$ per collision ( $1$ for H, $0.158$ for C).

Reactor problems: spent fuel stays radioactive for thousands of years and must be stored safely, and a loss of cooling can melt the core (Chernobyl 1986, Fukushima 2011). A reactor cannot explode like a bomb because its fuel is only slightly enriched.

Key idea

A reactor is a chain reaction held exactly at

k = 1

: the moderator makes neutrons slow enough to cause fission, the control rods remove the extra ones.

5. Nuclear Fusion

Nuclear fusion is the joining of two light nuclei into a heavier nucleus, with the release of energy because the product has a higher binding energy per nucleon.

_{1}^{2} H +_{1}^{2} H_{1}^{2} H +_{1}^{2} H_{1}^{2} H +_{1}^{3} H \to_{2}^{3} He + n + 3.27 MeV \to_{1}^{3} H +_{1}^{1} H + 4.03 MeV \to_{2}^{4} He + n + 17.59 MeV

The two D-D branches are about equally likely. D-T fusion releases the most energy and has the lowest ignition temperature, so it is the reaction used in fusion research.

Figure 9: D-T fusion,

^{2} H +^{3} H \to^{4} He + n + 17.6 MeV

. Equal and opposite momenta give

K \propto \frac{1}{m}

, so the neutron takes

\frac{4}{5}

Q

(14.0 MeV).

5.1 Why fusion needs very high temperature

Both nuclei are positive. To fuse, they must come within the range of the nuclear force, about $R_{1} + R_{2}$ , against Coulomb repulsion. The energy barrier is

U_{C} = \frac{k Z _{1} Z _{2} e ^{2}}{R _{1} + R _{2}} \approx \frac{1.44 \times 1 \times 1}{3.2} MeV \approx 0.44 MeV (D-T)

Figure 10: The Coulomb barrier for D-T fusion,

U = \frac{k e ^{2}}{r}

outside

R_{1} + R_{2} = R_{0} (2^{1/3} + 3^{1/3}) = 3.2 fm

. Its height 0.44 MeV would need about

3 \times 1 0^{9} K

if nuclei had to climb over it. A pair with energy

E

turns back at

r = \frac{k e ^{2}}{E}

(14.4 fm for the drawn

E = 0.1 MeV

); fast nuclei in the Maxwell tail tunnel through, so fusion starts near

1 0^{7} K

Setting $\frac{3}{2} k_{B} T = 0.44 MeV$ gives $T \approx 3 \times 1 0^{9} K$ . Fusion actually starts at $1 0^{7}$ to $1 0^{8} K$ for two reasons: a few nuclei in the high-energy tail of the Maxwell distribution are much faster than average, and quantum tunnelling lets them pass through the barrier without going over it. Because heat is what drives the nuclei together, fusion is also called a thermonuclear reaction. At such temperatures matter is a fully ionised gas called a plasma. High pressure alone cannot replace the temperature: it is the kinetic energy of the nuclei that must overcome the repulsion.

Exam Trick

Barrier energy in one line: $U = \frac{1.44 Z _{1} Z _{2}}{d ( fm )} MeV$ with $d$ the centre-to-centre distance at contact. Two deuterons of radius $2 fm$ touching ( $d = 4 fm$ ): $U = 0.36 MeV = 360 keV$ .

6. Energy of the Sun and Stars

Stars shine by fusing hydrogen into helium in their cores, where the Sun's temperature is about $1.5 \times 1 0^{7} K$ . In the Sun the main route is the proton-proton chain:

Figure 11: The proton-proton chain. Net:

4^{1} H \to^{4} He + 2 e^{+} + 2 ν + 26.7 MeV

, which equals

(4 m_{H} - m_{H e}) c^{2}

. The first step is a slow weak-interaction process, which is why the Sun burns steadily for billions of years.

4_{1}^{1} H \to_{2}^{4} He + 2 e^{+} + 2 ν + 26.7 MeV

About $0.7%$ of the mass of the hydrogen is converted into energy. The Sun radiates $3.8 \times 1 0^{26} W$ , so it loses about $4 \times 1 0^{9} kg$ of mass every second, yet has enough hydrogen for another five billion years.

Heavier stars also fuse through the CNO cycle, in which carbon acts as a catalyst. When the hydrogen in a star's core runs out, the core contracts and heats up until helium fuses into carbon ( $3^{4} He \to^{12} C$ ), and in massive stars the burning continues up to iron. Fusion stops at iron because the $\frac{B E}{A}$ curve peaks there. Elements heavier than iron are made by neutron capture, mostly in supernova explosions and neutron-star mergers.

6.1 Controlled thermonuclear fusion

A fusion reactor must hold a D-T plasma above about $1 0^{8} K$ at a high enough density for long enough. No material wall can touch such a plasma, so it is held by strong magnetic fields in a ring-shaped tokamak (the international ITER project, of which India is a member) or squeezed by powerful lasers (inertial confinement). Fusion fuel is plentiful (deuterium from sea water), and fusion produces little long-lived radioactive waste, but a practical power plant has not yet been built.

Figure 12: Per nucleon (that is, per kilogram of fuel) fusion beats fission by a factor of 4 to 8, and both beat chemical burning by about

1 0^{7}

. Per single reaction, however, fission (200 MeV) releases more than fusion (17.6 MeV).

Feature	Fission	Fusion
Process	Heavy nucleus splits	Light nuclei join
Typical energy per reaction	$\approx 200 MeV$	$3$ to $18 MeV$
Energy per nucleon (per kg of fuel)	$\approx 0.85 MeV$	$3.5 MeV$ (D-T), $6.7 MeV$ (p-p chain)
Condition needed	Slow neutrons, critical mass	Temperature $1 0^{7}$ to $1 0^{8} K$
Waste	Long-lived radioactive fragments	Mainly helium; little long-lived waste
Control	Achieved (power reactors)	Not yet achieved for power
Natural example	None today (Oklo, 2 billion years ago)	Sun and stars

Quick Recall: tap to check

Does fusion release more energy than fission?

Per kilogram of fuel, yes (about 4 to 8 times). Per single reaction, no:

17.6 MeV

against

200 MeV

Why is fusion called a thermonuclear reaction?

Only very high temperature gives the nuclei enough kinetic energy to approach against Coulomb repulsion.

What does the p-p chain convert, and how much energy per helium nucleus?

Four protons into one helium-4 nucleus, releasing

26.7 MeV

Key idea

Fission gives more energy per event, fusion more per kilogram. Fission is easy to start (a slow neutron) but leaves radioactive waste; fusion needs

1 0^{7} K

but powers every star.

7. Problem-Solving Map and Revision

Use the flowchart to choose a method, then the mind map to revise.

Figure 13: Solving fission and fusion problems. Most numericals are one of three types: Q from masses, Q from binding energies, or fuel needed for a given power.

Figure 14: Mind map of this concept. Read the left column, then the right; cover a branch, recall its three points, then check.

8. Solved Examples

Solved Example 1

Find the energy released in the fission

_{92}^{235} U +_{0}^{1} n \to_{56}^{141} Ba +_{36}^{92} Kr + 3_{0}^{1} n

. Atomic masses:

^{235} U = 235.043928 u

^{141} Ba = 140.914404 u

^{92} Kr = 91.926173 u

m_{n} = 1.008665 u

Solution:

With nuclear masses, initial mass $= M_{U} - 92 m_{e} + m_{n}$ and final mass $= (M_{B a} - 56 m_{e}) + (M_{K r} - 36 m_{e}) + 3 m_{n}$ . The $92 m_{e}$ cancel, leaving atomic masses:

Q = [(M_{U} + m_{n}) - (M_{B a} + M_{K r} + 3 m_{n})] c^{2}

Initial: $235.043928 + 1.008665 = 236.052593 u$ . Final: $140.914404 + 91.926173 + 3.025995 = 235.866572 u$ .

$Δ m = 0.186021 u$ ; $Q = 0.186021 \times 931.5$ .

Answer: $Q \approx 173 MeV$ for this channel (about $200 MeV$ once the fragments' later decays are included).

Solved Example 2

Find the Q values of the two D-D reactions (a)

^{2} H +^{2} H \to^{3} He + n

and (b)

^{2} H +^{2} H \to^{3} H +^{1} H

. Atomic masses:

^{2} H = 2.014102

^{3} He = 3.016029

^{3} H = 3.016049

^{1} H = 1.007825

n = 1.008665 u

Solution:

(a) Nuclear masses: initial $2 (M_{D} - m_{e})$ , final $(M_{H e} - 2 m_{e}) + m_{n}$ . The electrons cancel: $Q = [2 M_{D} - M_{H e} - m_{n}] c^{2}$

$= (4.028204 - 3.016029 - 1.008665) \times 931.5 = 0.003510 \times 931.5$ .

(b) $Q = [2 M_{D} - (M_{T} + M_{H})] c^{2} = (4.028204 - 4.023874) \times 931.5 = 0.004330 \times 931.5$ .

Answer: (a) $3.27 MeV$ ; (b) $4.03 MeV$ .

Solved Example 3

In the D-T reaction

^{2} H +^{3} H \to^{4} He + n

, find

Q

and the kinetic energy of the neutron (initial kinetic energies negligible).

^{3} H = 3.016049 u

^{4} He = 4.002603 u

Solution:

$Q = [(M_{D} + M_{T}) - (M_{H e} + m_{n})] c^{2} = (5.030151 - 5.011268) \times 931.5 = 0.018883 \times 931.5 = 17.59 MeV$ .

Momentum is conserved, so $p_{H e} = p_{n}$ and $K = \frac{p ^{2}}{2 m}$ gives $\frac{K _{n}}{K _{H e}} = \frac{m _{H e}}{m _{n}} \approx 4$ .

$K_{n} = \frac{4}{5} Q$ .

Answer: $Q = 17.6 MeV$ , $K_{n} \approx 14.1 MeV$ (and $K_{H e} \approx 3.5 MeV$ ).

Solved Example 4

A nuclear power station produces

1000 MW

from

^{235} U

, with

200 MeV

per fission. Assuming

100%

efficiency, how much

^{235} U

is used in one year?

Solution:

Energy per fission $= 200 \times 1.6 \times 1 0^{- 13} = 3.2 \times 1 0^{- 11} J$ .

Fissions per second $= \frac{1 0 ^{9}}{3.2 \times 1 0 ^{- 11}} = 3.1 \times 1 0^{19} s^{- 1}$ .

In a year: $3.1 \times 1 0^{19} \times 3.15 \times 1 0^{7} = 9.9 \times 1 0^{26}$ nuclei.

Mass $= \frac{9.9 \times 1 0 ^{26}}{6.02 \times 1 0 ^{23}} \times 235 g = 3.85 \times 1 0^{5} g$ .

Answer: about $385 kg$ of $^{235} U$ per year. Coal giving the same heat would weigh about a million tonnes.

Solved Example 5

Two deuterons, each of radius

2.0 fm

, must touch for fusion. Find the height of the Coulomb barrier, and the temperature at which the average thermal energy of a deuteron equals it.

Solution:

At contact the centres are $d = 4.0 fm$ apart. $U = \frac{k e ^{2}}{d} = \frac{1.44 MeV fm}{4.0 fm} = 0.36 MeV$ .

Each deuteron has average thermal energy $\frac{3}{2} k_{B} T$ , and the pair together must supply $U$ : $2 \times \frac{3}{2} k_{B} T = U$ .

$T = \frac{U}{3 k _{B}} = \frac{0.36 \times 1.6 \times 1 0 ^{- 13}}{3 \times 1.38 \times 1 0 ^{- 23}} = 1.4 \times 1 0^{9} K$ .

Answer: barrier $\approx 360 keV$ ; $T \sim 1 0^{9} K$ . Actual fusion needs only about $1 0^{7} K$ thanks to tunnelling and the Maxwell tail.

Solved Example 6

The Sun radiates

3.8 \times 1 0^{26} W

. Find (a) the mass it loses per second and (b) the number of protons it fuses per second, given

26.7 MeV

per helium nucleus formed.

Solution:

(a) $\frac{Δ m}{Δ t} = \frac{P}{c ^{2}} = \frac{3.8 \times 1 0 ^{26}}{9 \times 1 0 ^{16}} = 4.2 \times 1 0^{9} kg s^{- 1}$ .

(b) Helium nuclei per second $= \frac{3.8 \times 1 0 ^{26}}{26.7 \times 1.6 \times 1 0 ^{- 13}} = 8.9 \times 1 0^{37}$ ; each uses 4 protons.

Answer: (a) $4.2 \times 1 0^{9} kg$ each second; (b) $3.6 \times 1 0^{38}$ protons per second.

Solved Example 7

The function of the moderator in a nuclear reactor is to
(A) absorb neutrons
(B) slow down neutrons
(C) cool the core
(D) stop gamma rays

Solution:

Answer: (B). The moderator slows the fast fission neutrons to thermal energies, at which $^{235} U$ captures them readily. Absorbing neutrons is the job of control rods, carrying heat that of the coolant, and stopping radiation that of the shield.

Solved Example 8

Which statement is correct?
(A) Fusion releases more energy per reaction than fission
(B) Fusion releases more energy per unit mass of fuel than fission
(C) Fission needs a temperature of

1 0^{7} K

(D) The nuclear force acts only between protons

Solution:

Answer: (B). D-T fusion gives $3.5 MeV$ per nucleon against $0.85 MeV$ for fission, so more energy per kilogram, although $17.6 MeV$ per reaction is less than $200 MeV$ per fission. It is fusion that needs $1 0^{7} K$ , and the nuclear force acts equally between all nucleons.

Solved Example 9

In a chain reaction the multiplication factor is

k = 1.01

and the time between generations is

1 ms

. By what factor does the power grow in

1 s

Solution:

Generations in $1 s$ : $n = \frac{1}{1 0 ^{- 3}} = 1000$ .

Growth factor $= k^{n} = 1.0 1^{1000} = e^{1000 l n 1.01} = e^{9.95}$ .

Answer: about $2.1 \times 1 0^{4}$ times. Even $1%$ above critical is dangerous; reactors are controlled only because some neutrons are delayed by seconds.

Solved Example 10

A neutron makes a head-on elastic collision with a deuteron at rest. The fraction of its kinetic energy it loses is
(A)

1/9

(B)

8/9

(C)

1/3

(D)

2/3

Solution:

Answer: (B). With $A = 2$ : $\frac{Δ K}{K} = \frac{4 A}{( A + 1 ) ^{2}} = \frac{8}{9}$ . The neutron bounces back with $\frac{1}{3}$ of its speed, so it keeps $\frac{1}{9}$ of its energy: option (A) is the fraction kept, (C) the speed ratio.

Solved Example 11

In each head-on collision with a carbon-12 nucleus a neutron keeps

(\frac{11}{13})^{2}

of its kinetic energy. How many head-on collisions would slow a

2 MeV

neutron to

0.025 eV

Solution:

After $n$ collisions $K_{n} = K_{0} (\frac{11}{13})^{2 n}$ . Set $\frac{K _{n}}{K _{0}} = \frac{0.025}{2 \times 1 0 ^{6}} = 1.25 \times 1 0^{- 8}$ .

$2 n ln \frac{13}{11} = ln (8 \times 1 0^{7})$ , so $2 n \times 0.1671 = 18.20$ and $n = 54.5$ .

Answer: about 55 head-on collisions. Real collisions are mostly glancing, so the average is about $115$ (Figure 8); a proton-rich moderator such as water needs only about $18$ .

Practice Questions

Find the Coulomb force between two protons $2 fm$ apart.Answer: $57.5 N$
Find the energy released when $1 g$ of $^{235} U$ undergoes fission ( $200 MeV$ per fission).Answer: $8.2 \times 1 0^{10} J$
How many fissions per second produce $1 W$ ( $200 MeV$ per fission)?Answer: $3.1 \times 1 0^{10}$
Find the height of the Coulomb barrier for two protons that must come within $2 fm$ .Answer: $0.72 MeV$
In $^{235} U + n \to^{140} Xe +^{94} Sr + x n$ , find $x$ .Answer: $2$
Find the energy released per kg of D-T fuel ( $17.6 MeV$ per reaction, $5 u$ of fuel per reaction).Answer: $3.4 \times 1 0^{14} J$
Neutrons of $2 MeV$ are slowed to about $0.025 eV$ . By roughly what factor is their speed reduced?Answer: About $9000$ ( $8 \times 1 0^{7}$ )
A neutron collides head-on and elastically with a nucleus of mass number $A$ at rest and keeps $\frac{1}{4}$ of its kinetic energy. Find $A$ .Answer: $A = 3$ ( $\frac{A - 1}{A + 1} = \frac{1}{2}$ )

Common Mistakes to Avoid

Watch out

Saying the nuclear force acts only between protons, or depends on charge. It is charge independent: $n$ - $n$ , $n$ - $p$ and $p$ - $p$ forces are nearly equal.
Calling the nuclear force always attractive. It becomes strongly repulsive below about $0.8 fm$ .
Thinking fast neutrons are best for fission of $^{235} U$ . Slow (thermal) neutrons are captured far more readily; that is why a moderator is needed.
Confusing moderator and control rods. The moderator slows neutrons; control rods absorb them.
Claiming fusion releases more energy per reaction than fission. It releases more per nucleon (per kg), not per event.
Believing high pressure can replace high temperature in fusion. Nuclei need kinetic energy to approach against Coulomb repulsion.
Forgetting the free neutrons when balancing fission: check both $A$ and $Z$ on each side ( $235 + 1 = 141 + 92 + 3$ ).
Sharing the Q value equally between two products. At rest, momenta are equal, so $K \propto \frac{1}{m}$ : the lighter particle takes more.

Frequently Asked Questions

What are the properties of nuclear force?

The nuclear force is the strongest force in nature, about 100 times the Coulomb force inside a nucleus. It is short-range, acting over only about 2 to 3 femtometre, charge independent, saturating, strongly repulsive below about 0.8 femtometre, spin dependent and non-central.

What is nuclear fission?

Nuclear fission is the splitting of a heavy nucleus such as uranium-235 into two middle-mass fragments after it absorbs a slow neutron. Two or three new neutrons and about 200 MeV of energy are released, mostly as kinetic energy of the fragments.

What is a chain reaction and how is it controlled?

Neutrons released in one fission cause further fissions, so the reaction sustains itself. In a reactor, cadmium or boron control rods absorb extra neutrons so that exactly one neutron from each fission causes another, keeping the multiplication factor equal to one.

Why is a moderator used in a nuclear reactor?

Fission releases fast neutrons of about 2 MeV, but uranium-235 captures slow neutrons far more readily. A moderator of light nuclei such as water, heavy water or graphite slows the neutrons by collisions to thermal energies of about 0.025 eV.

Why does nuclear fusion need such high temperatures?

The nuclei are positively charged and repel each other. They must come within the range of the nuclear force, which means overcoming a Coulomb barrier of a few hundred keV. Only at temperatures of ten million kelvin or more do enough nuclei move fast enough to tunnel through.

What is the source of energy of the Sun?

The Sun shines by nuclear fusion in its core at about 15 million kelvin. Through the proton-proton chain, four hydrogen nuclei become one helium nucleus, releasing 26.7 MeV. The Sun converts about four billion kilograms of mass into energy every second.

Which nuclear energy questions come in NEET?

NEET asks properties of nuclear force, the energy released in fission from masses or binding energy per nucleon, the role of moderator and control rods, the fusion reaction in the Sun and the comparison of fission and fusion. Remember about 200 MeV per fission and 931.5 MeV per u.

How are fission and fusion tested in JEE Main and Advanced?

JEE Main tests Q values from atomic masses, energy per fission, fuel consumption of a reactor and the Coulomb barrier for fusion. JEE Advanced adds sharing of kinetic energy between products by momentum conservation, chain-reaction growth and energy balance in stars.

Previous year questions on Nuclear Forces and Nuclear Energy

3 questions from past papers, each with a step-by-step solution.

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