**CBSE NOTES CLASS 12 PHYSICS**

**CHAPTER 13 ****NUCLEUS**

**Subatomic particles**

Nucleon |
Mass |
Charge |

Proton |
1.6727 x 10 Or 1.00727 u |
1.602×10 |

Neutron |
1.6750 x 10 Or 1.00866 u |
Nil |

Electron |
9.110 x 10 Or 0.00055 u |
- 1.602×10 |

**Nucleus**

Rutherford proposed that the entire positive charge and nearly the entire mass of atom are concentrated in a very small space called the nucleus of an atom.

The nucleus consists of protons and neutrons. They are called **nucleons**.

Z - atomic number = number of protons

N - neutron number = number of neutrons

A - mass number = Z + N = total number of protons and neutrons

Nuclear species or nuclides are shown by the notation ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$ where X is the chemical symbol of the species. For example, the nucleus of gold is denoted by${}_{79\mathrm{}}{}^{197}\mathrm{A}\mathrm{u}$. It contains 197 nucleons, of which 79 are protons and the rest 118 are neutrons.

**Atomic mass unit**

It is defined as 1/12^{th} the mass of carbon nucleus. It is abbreviated as amu and often denoted by u. Thus

$$1\mathrm{}\mathrm{u}\mathrm{}=\mathrm{}\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{}{}_{\mathrm{}}{}^{12}\mathrm{C}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}}{12}$$

$$=\mathrm{}\frac{1.992647\mathrm{}\times {10}^{26}\mathrm{}\mathrm{}\mathrm{k}\mathrm{g}}{12}$$

$$=\mathrm{}1.660539\mathrm{}\times {10}^{-27}\mathrm{}\mathrm{k}\mathrm{g}$$

**Mass energy**

Mass is a form of energy and can be converted into other forms of energy, as follows

E = mc^{2}

**Mass defect **

The difference between the sum of masses of all nucleons (m) and mass of the nucleus (M) is called mass defect.

Mass defect (ΔM) = m – M

= [Zm_{p} + (A – Z)m_{n} – M]

**Nuclear force **

The force acting inside the nucleus or acting between nucleons is called nuclear force. Nuclear force is the strongest forces in nature. It is

- very short range attractive force.
- non-central, non-conservative force.
- neither gravitational nor electrostatic force.
- independent of charge.
- 100 times that of electrostatic force and 10
^{38}times that of gravitational force.

**Nuclear binding energy **

The minimum energy required to separate the nucleons up to an infinite distance from the nucleus, is called nuclear binding energy.

$${\mathrm{E}}_{\mathrm{b}}\mathrm{}=\mathrm{}\mathrm{\Delta}\mathrm{M}{\mathrm{c}}^{2}$$

**Binding energy per nucleon**

$${\mathrm{E}}_{\mathrm{b}\mathrm{n}}\mathrm{}=\frac{{\mathrm{E}}_{\mathrm{b}}}{\mathrm{A}}$$

- The binding energy per nucleon, E
_{bn}, is practically constant, i.e. practically independent of the atomic number for nuclei of middle mass number (30 < A < 170). - It is maximum, about 8.75 MeV, for A = 56 and has a value of 7.6 MeV for A = 238.
- E
_{bn}is lower for both light nuclei (A < 30) and heavy nuclei (A > 170).

**Properties of binding force and binding energy**

- The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.
- The constancy of the binding energy in the range 30 < A < 170 is due to the fact that the nuclear force is short-ranged. Let us consider a particular nucleon inside a sufficiently large nucleus. It will be under the influence of only some of its neighbours, which come within the range of the nuclear force. If any other nucleon is at a distance more than the range of the nuclear force from the particular nucleon it will have no influence on the binding energy of the nucleon under consideration. If a nucleon can have a maximum of
**p****neighbours**within the range of nuclear force, its binding energy would be proportional to**p**. Let the binding energy of the nucleus be**pk,**where k is a constant having the dimensions of energy. If we increase A by adding nucleons they will not change the binding energy of a nucleon inside. Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. The binding energy per nucleon is a constant and is approximately equal to**pk**.The property that a given nucleon influences only nucleons close to it is referred to as

**saturation property**of the nuclear force. - A very heavy nucleus, say A = 240, has lower binding energy per nucleon compared to that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightly bound. This implies energy would be released in the process. This is the principle of energy production through
**fission.** - If we consider two very light nuclei (A ≈ 10) joining to form a heavier nucleus. The binding energy per nucleon of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of
**fusion**.

_{o}

_{o}

**Relation of nuclear size and atomic mass number**

The radius of the nucleus R ∝ A^{1/3}

$$\Rightarrow \mathrm{}\mathrm{R}\mathrm{}=\mathrm{}{\mathrm{R}}_{\mathrm{o}}{\mathrm{A}}^{1/3}\mathrm{}$$

where, R_{o} = 1.2 × 10^{-15} m is an empirical constant.

**Nuclear density**

Nuclear density is independent of mass number and therefore same for all nuclei.

$$\mathrm{N}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{\rho}\mathrm{}=\mathrm{}\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s}}{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s}}\mathrm{}$$

$$\Rightarrow \mathrm{\rho}=\frac{3\mathrm{A}\mathrm{m}}{4\mathrm{\pi}{{\mathrm{A}\mathrm{R}}_{\mathrm{o}}}^{3}}=\mathrm{}\frac{3\mathrm{m}}{4\mathrm{\pi}{{\mathrm{R}}_{\mathrm{o}}}^{3}}$$

where, m = average mass of a nucleon.

**Isotopes**

The atoms of an element having same atomic number but different mass numbers are called isotopes.

e.g., ^{1}_{1}H, ^{2}_{1}H, ^{3}_{1}H are isotopes of hydrogen.

**Isobars**

The atoms of different elements having same mass numbers but different atomic numbers are called isobars.

**Isotones**

The atoms of different elements having different atomic numbers and different mass numbers but having same number of neutrons are called isotones.

**Isomers**

Atoms having the same mass number and the same atomic number but different radioactive properties are called isomers,

**Radioactivity**

The phenomena of disintegration of heavy elements into comparatively lighter elements by the emission of radiations is called radioactivity.

Three types of radiations are emitted by radioactive elements

**α-rays**

α-rays consists of α-particles, which are doubly ionised helium ion.

When an α-particle is emitted by a nucleus its atomic number decreases by 2 and mass number decreases by 4.

${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}\mathrm{}\mathrm{}{}_{\mathrm{Z}-2}{}^{\mathrm{A}-4}\mathrm{Y}\mathrm{}+{}_{2}{}^{4}\mathrm{H}\mathrm{e}$

Example

${}_{92}{}^{238}\mathrm{U}\mathrm{}\mathrm{}{}_{90}{}^{234}\mathrm{T}\mathrm{h}\mathrm{}+{}_{2}{}^{4}\mathrm{H}\mathrm{e}$

Total mass of RHS is less than that of LHS.

The disintegration energy or the **Q-value of a nuclear reaction** is the difference between the initial mass energy and the total mass energy of the decay products. For α-decay

$$\mathrm{Q}\mathrm{}=\mathrm{}[{\mathrm{m}}_{\mathrm{X}}\mathrm{}\u2013\mathrm{}({\mathrm{m}}_{\mathrm{Y}}\mathrm{}+\mathrm{}{\mathrm{m}}_{\mathrm{H}\mathrm{e}}\left)\right]\mathrm{}{\mathrm{c}}^{2}$$

Q is also the net kinetic energy gained in the process or, if the initial nucleus X is at rest, the kinetic energy of the products, Q > 0 for exothermic processes such as α-decay.

**β-rays**

β-rays consist of fast moving electrons.

When a β-particle is emitted by a nucleus its atomic number is increases by 1 and mass number remains unchanged.

The emission of electron in β^{−} decay is accompanied by the emission of an antineutrino ($\stackrel{\u203e}{\nu}$); in β^{+} (positron) decay a neutrino (ν) is generated. Neutrinos are neutral particles with very small (possiblly, even zero) mass compared to electrons. They have only weak interaction with other particles. They are, therefore, very difficult to detect, since they can penetrate large quantity of matter (even earth) without any interaction.

$\mathrm{n}\mathrm{}\mathrm{}\mathrm{p}+{\mathrm{e}}^{-}+\stackrel{\u203e}{\mathrm{\nu}};\mathrm{}{\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{\beta}}^{-}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}\mathrm{}$

$$\mathrm{p}\mathrm{}\mathrm{}\mathrm{n}+{\mathrm{e}}^{+}+\mathrm{\nu};{\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{\beta}}^{+}\mathrm{}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}$$

${}_{15}{}^{32}\mathrm{P}\mathrm{}\mathrm{}{}_{16}{}^{32}\mathrm{S}\mathrm{}+{\mathrm{e}}^{-}+\stackrel{\u203e}{\mathrm{\nu}}\mathrm{};\mathrm{}{\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{\beta}}^{-}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}\mathrm{}$

${}_{11}{}^{22}\mathrm{N}\mathrm{a}\mathrm{}\mathrm{}{}_{10}{}^{22}\mathrm{N}\mathrm{e}\mathrm{}+{\mathrm{e}}^{+}+\mathrm{\nu};{\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{\beta}}^{+}\mathrm{}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}\mathrm{}$

**γ-rays**

γ-rays are electromagnetic rays.

When a γ-particle is emitted by a nucleus its atomic number and mass number remain unchanged.

When a nucleus in an excited state spontaneously decays to its ground state (or to a lower energy state), a photon is emitted with energy equal to the difference in the two energy levels of the nucleus. This is called gamma decay.

A gamma ray is emitted when an α or β decay results in a daughter nucleus in an excited state, which returns to the ground state by a single photon transition or successive transitions involving more than one photon. For example

$${}_{27}{}^{60}\mathrm{C}\mathrm{o}\mathrm{}\mathrm{}\mathrm{}{}_{28}{}^{60}\mathrm{N}\mathrm{i}+{\mathrm{e}}^{-}+\stackrel{\u203e}{\mathrm{\nu}};\mathrm{}\mathrm{}$$

Ni is in excited state, which results in successive emission of gamma rays of energies 1.17 MeV and 1.33 MeV.

**Law of radioactive decay **

The rate of disintegration of radioactive atoms at any instant is directly proportional to the number of radioactive atoms present in the sample at that instant.

Rate of disintegration

$$\left(-\frac{\mathrm{d}\mathrm{N}}{\mathrm{d}\mathrm{t}}\right)\propto \mathrm{}\mathrm{N}\mathrm{}$$

$$\Rightarrow -\frac{\mathrm{d}\mathrm{N}}{\mathrm{d}\mathrm{t}}=\mathrm{\lambda}\mathrm{N},$$

where λ is called the radioactive decay constant or disintegration constant

$$\Rightarrow -\frac{\mathrm{d}\mathrm{N}}{\mathrm{N}}=\mathrm{\lambda}\mathrm{d}\mathrm{t}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}{\int}_{{\mathrm{N}}_{0}}^{\mathrm{N}}\frac{\mathrm{d}\mathrm{N}}{\mathrm{N}}=\mathrm{}{\int}_{{\mathrm{t}}_{0}}^{\mathrm{t}}\mathrm{\lambda}\mathrm{d}\mathrm{t}\mathrm{}$$

$$\Rightarrow \mathrm{ln}\mathrm{N}-\mathrm{ln}{\mathrm{N}}_{0}\mathrm{}=\mathrm{}\mathrm{\lambda}\left(\mathrm{t}-{\mathrm{t}}_{0}\right)\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{N}\mathrm{}=\mathrm{}{\mathrm{N}}_{0}{\mathrm{e}}^{-\mathrm{\lambda}\mathrm{t}}\mathrm{}\mathrm{}\mathrm{}$$

Therefore, the number of atoms present undecayed in the sample at any instant,

$$\mathrm{N}\mathrm{}=\mathrm{}{\mathrm{N}}_{0}{\mathrm{e}}^{-\mathrm{\lambda}\mathrm{t}}$$

where, N_{o} is number of atoms at time t = 0 and N is number of atoms at time t.

**Rate of decay **

The total decay rate R of a sample is the number of nuclei disintegrating per unit time.

$$\mathrm{R}\mathrm{}=\mathrm{}\u2013\frac{\mathrm{d}\mathrm{N}}{\mathrm{d}\mathrm{t}}$$

$$\Rightarrow \mathrm{}\mathrm{R}\mathrm{}=\mathrm{}\mathrm{\lambda}{\mathrm{N}}_{\mathrm{o}}{\mathrm{e}}^{-\mathrm{\lambda}\mathrm{t}}=\mathrm{}{\mathrm{R}}_{\mathrm{o}}{\mathrm{e}}^{-\mathrm{\lambda}\mathrm{t}}$$

$$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{}{\mathrm{R}}_{\mathrm{o}}\mathrm{}=\mathrm{}\mathrm{\lambda}{\mathrm{N}}_{\mathrm{o}}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}\mathrm{}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{}\mathrm{t}\mathrm{}=\mathrm{}0.\mathrm{}$$

We can therefore, write,

$$\mathrm{R}\mathrm{}=\mathrm{}\mathrm{\lambda}\mathrm{N}$$

The decay rate of a sample is called **activity**. The SI unit for activity is **Becquerel**

Another unit of activity is curie.

1 curie = 1 Ci = 3.7×10^{10 }decays per second = 3.7×10^{10} Bq

**Half-life of a radioactive element**

The time in which half of the number of atoms present initially in any sample get decayed; is called half-life (T_{1/2}) of that radioactive element.

Relation between half-life and disintegration constant is given by

$${\mathrm{T}}_{\frac{1}{2}}\mathrm{}=\frac{\mathrm{ln}2}{\mathrm{\lambda}}=\frac{0.6931}{\mathrm{\lambda}}$$

**Average life or mean life of a radioactive element(τ)**

Average life or mean life (τ) of a radioactive element is the ratio of total life time of all the atoms and total number of atoms present initially in the sample.

$$\tau =\frac{1}{\lambda}=1.44{T}_{\frac{1}{2}}$$

The number of atoms left un-decayed after ‘n’ half-lives is given by

$$\mathrm{N}\mathrm{}=\mathrm{}{\mathrm{N}}_{\mathrm{o}}{\left(\frac{1}{2}\right)}^{\mathrm{n}}\mathrm{}=\mathrm{}{\mathrm{N}}_{\mathrm{o}}{\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{{T}_{\frac{1}{2}}}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

$$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e},\mathrm{}\mathrm{}\mathrm{n}\mathrm{}=\mathrm{}\frac{\mathrm{t}}{{T}_{\frac{1}{2}}},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{}\mathrm{t}\mathrm{}=\mathrm{}\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}$$

**Nuclear fission**

When a heavy nucleus decays into two or more intermediate mass fragments the reaction (activity) is called fission.

${}_{92}{}^{235}\mathrm{U}+{}_{0}{}^{1}\mathrm{n}\mathrm{}{}_{92}{}^{236}\mathrm{U}\mathrm{}{}_{46}{}^{144}\mathrm{B}\mathrm{a}+{}_{39}{}^{89}\mathrm{K}\mathrm{r}+3\mathrm{}{}_{0}{}^{1}\mathrm{n}\mathrm{}$

$${}_{92}{}^{235}\mathrm{U}+{}_{0}{}^{1}\mathrm{n}\mathrm{}{}_{92}{}^{236}\mathrm{U}\mathrm{}{}_{51}{}^{133}\mathrm{S}\mathrm{b}+{}_{41}{}^{99}\mathrm{N}\mathrm{b}+4\mathrm{}{}_{0}{}^{1}\mathrm{n}$$

$${}_{92}{}^{235}\mathrm{U}+{}_{0}{}^{1}\mathrm{n}\mathrm{}{}_{92}{}^{236}\mathrm{U}\mathrm{}{}_{54}{}^{140}\mathrm{X}\mathrm{e}+{}_{38}{}^{94}\mathrm{S}\mathrm{r}+2\mathrm{}{}_{0}{}^{1}\mathrm{n}$$

**Structure of nuclear reactor**

**Nuclear fuel**

Usually pellets of uranium oxide (UO_{2}) are arranged in tubes to form fuel rods. The rods are arranged into fuel assemblies in the reactor core.

**Moderator**

Material in the core which slows down the neutrons released from fission so that they cause more fission. It is usually water, but may be heavy water or graphite.

**Control rods**

These are made with neutron-absorbing material such as cadmium, hafnium or boron, and are inserted or withdrawn from the core to control the rate of reaction, or to halt it

**Coolant in nuclear reactor**

This is a fluid circulating through the core so as to transfer the heat from it. In light water reactors the water moderator functions also as primary coolant.

**Pressure vessel or pressure tubes in nuclear reactor**

Usually a robust steel vessel containing the reactor core and moderator/ coolant, but it may be a series of tubes holding the fuel and conveying the coolant through the surrounding moderator.

**Steam generator in nuclear generator**

This is the part of the cooling system of pressurised water reactors, where the high-pressure primary coolant bringing heat from the reactor is used to make steam for the turbine.

**Nuclear fusion **

When light nuclei fuse into a heavier nucleus it is called fusion. Energy in stars is due to fusion.

${}_{1}{}^{1}H+{}_{1}{}^{1}\mathrm{H}\mathrm{}{}_{1}{}^{2}\mathrm{H}\mathrm{e}+\mathrm{\nu}+0.42\mathrm{}\mathrm{M}\mathrm{e}\mathrm{V}\mathrm{}$

${}_{1}{}^{2}\mathrm{H}+{}_{1}{}^{2}\mathrm{H}\mathrm{}{}_{2}{}^{3}\mathrm{H}\mathrm{e}+\mathrm{n}+3.27\mathrm{}\mathrm{M}\mathrm{e}\mathrm{V}\mathrm{}$

$${}_{1}{}^{2}\mathrm{H}+{}_{1}{}^{2}\mathrm{H}\mathrm{}{}_{1}{}^{3}\mathrm{H}\mathrm{e}+{}_{1}{}^{1}\mathrm{H}+4.03\mathrm{}\mathrm{M}\mathrm{e}\mathrm{V}$$

For fusion to take place, the two nuclei must come close enough so that attractive short-range nuclear force is able to affect them. However, since they are both positively charged particles, they experience coulomb repulsion. They, therefore, must have enough energy to overcome this coulomb barrier. The height of the barrier depends on the charges and radii of the two interacting nuclei.

The temperature at which two protons in a proton gas would (on an average) have enough energy to overcome the coulomb barrier:

(3/2)kT = KE ≈ 400 keV, which gives T ~ 3 × 10^{9} K.

When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the coulomb repulsive behaviour, it is called **thermonuclear fusion**.

**The proton-proton (p, p) cycle in the sun**

${}_{1}{}^{1}\mathrm{H}+{}_{1}{}^{1}\mathrm{H}\mathrm{}\mathrm{}{}_{1}{}^{2}\mathrm{H}\mathrm{e}+\mathrm{\nu}+0.42\mathrm{}\mathrm{M}\mathrm{e}\mathrm{V}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\left(\mathrm{i}\right)$

${\mathrm{e}}^{-}+{\mathrm{e}}^{-}\mathrm{}\mathrm{}\mathrm{\gamma}+\mathrm{\gamma}+1.02\mathrm{}\mathrm{M}\mathrm{e}\mathrm{V}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\left(\mathrm{i}\mathrm{i}\right)$

${}_{1}{}^{2}\mathrm{H}+{}_{1}{}^{1}\mathrm{H}\mathrm{}\mathrm{}{}_{2}{}^{3}\mathrm{H}\mathrm{e}+\mathrm{\gamma}+5.49\mathrm{}\mathrm{M}\mathrm{e}\mathrm{V}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\mathrm{}$

$${}_{2}{}^{3}\mathrm{H}\mathrm{e}+{}_{2}{}^{3}\mathrm{H}\mathrm{e}\mathrm{}\mathrm{}{}_{2}{}^{4}\mathrm{H}\mathrm{e}+{}_{1}{}^{1}\mathrm{H}+{}_{1}{}^{1}\mathrm{H}+12.86\mathrm{}\mathrm{M}\mathrm{e}\mathrm{V}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\left(\mathrm{i}\mathrm{v}\right)$$

The final balanced reaction after combining (i) to (iv)

$4\mathrm{}{}_{1}{}^{1}\mathrm{H}+2{\mathrm{e}}^{-}\mathrm{}{}_{2}{}^{4}\mathrm{H}\mathrm{e}+2\mathrm{\nu}+6\mathrm{\gamma}+26.7\mathrm{}\mathrm{M}\mathrm{e}\mathrm{V}\mathrm{}$ or

$$\left(4\mathrm{}{}_{1}{}^{1}\mathrm{H}+4{\mathrm{e}}^{-}\right)\mathrm{}\left({}_{2}{}^{4}\mathrm{H}\mathrm{e}+2{\mathrm{e}}^{-}\right)+2\mathrm{\nu}+6\mathrm{\gamma}+26.7\mathrm{}\mathrm{M}\mathrm{e}\mathrm{V}$$

As the hydrogen in the core gets depleted and becomes helium, the core starts to cool. The star begins to collapse under its own gravity which increases the temperature of the core. If this temperature increases to about 10^{8} K, fusion takes place again, this time of helium nuclei into carbon. This process can generate through fusion higher and higher mass number elements. But elements more massive than those near the peak of the binding energy curve in cannot be produced.

The age of the sun is about 5×10^{9} y and it is estimated that there is enough hydrogen in the sun to keep it going for another 5 billion years.

After that, the hydrogen burning will stop and the sun will begin to cool and will start to collapse under gravity, which will raise the core temperature. The outer envelope of the sun will expand, turning it into the so called **red giant**.

Controlled thermonuclear reaction may be possible in future.