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Nuclei

nucleus of an α-particle of kinetic energy 5.5 MeV is about 4.0 × 10–14 m. ... Here the energy equivalent of mass m is related by the above equation ... been achieved in the study of nuclear reactions amongst nucleons, nuclei, ... It is here that Einstein's ... Unlike Coulomb's law or the Newton's law of gravitation there is no.

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Lawson criterion - Wikipedia

The Lawson criterion is a figure of merit used in nuclear fusion research. It compares the rate of energy being generated by fusion reactions within the fusion fuel ...

For more information, see Lawson criterion - Wikipedia

Required formulas:

The amount of energy consumed or emitted during a nuclear reaction is known as the Q value. I can be determined from the difference between the masses of the initial reactants and the masses of the final products of a nuclear reaction. (MeV is a unit of energy).

Coulomb's law states that the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

F = \frac{1}{4 \pi \epsilon_0} * \frac{q_1q_2 }{d^2} , where \text{ constant } \frac{1}{4 \pi \epsilon_0} = 9 * 10^{9} Nm^2/C^2 ,   q_1 \text{ and } q_2 -  charges and d -  distance between the two charges

Potential energy of two nuclei undergoing fusion P.E = \frac{1}{4 \pi \epsilon_0} \frac{q^2}{d}

Given that

deuterium–tritium fusion: ^{2}_{1} H + ^{3}_{1} H \rightarrow ^{4}_{2} He + n   

m (^{2}_{1} H) = 2.014102 u

m (^{3}_{1} H) = 3.016049 u

mass of neutron ( m_n ) =  1.00867

Mass of ^{4}_{2} He = 4.0026

a) Energy released

Q - value = \Delta m c^2   

\Delta m = \text{ mass of } ^{2}_{1} H + \text{ mass of } ^{3}_{1} H - \text{ mass of } ^{4}_{2} He - m_n

[math] \Delta m = [2.014102 + 3.016049 - 4.0026 - 1.00867] u [/math]

Q - value = 0.01888 c^2 * 931.5 MeV/c^2

Q-value = 17.59 MeV

b)

Charge on deuterium and tritium q = 1.6 * 10^{- 19} C  

radius of deuterium and tritium r = 2 femtometre (fm) = 2 * 10^{-15} m   

Distance between the two charges d = radius of deuterium + radius of tritium = 2r

The amount of kinetic energy needed to overcome Coulomb repulsion is equal to the amount of potential energy. (K.E = P.E)

Potential energy P.E = \frac{1}{4 \pi \epsilon_0} \frac{(1.6 * 10^{- 19})^2}{2 * (2 * 10^{-15})}

P.E = \frac{9 * 10^9 (1.6 * 10^{- 19})^2}{4*10^{-15}}

P.E = 5.76 * 10^{-14} Joules

K.E = 5.76 * 10^{-14} joules

The kinetic energy of all gas molecules is a measure of the temperature

Kinetic temperature = \frac{3}{2} kT where, k - Boltzmann constant, T- temperature.

K.E = \frac{3}{2} * 1.38 * 10^{-23} * T   

5.76 * 10^{-14} = \frac{3}{2} * 1.38 * 10^{-23} * T

T = 2.78 * 10^9 K

Hence, the temperature needed to start the reaction T = 2.78 * 10^9 K