In the reaction X Z A → Y Z - 2 A - 4 + H 2 4 e + Q of the nucleus X at rest. Taking the ratio of the mass of the α - particle M u and the mass of Y nucleus as M U M Y = 4 A - 4 show that the Q value of the reaction is given by Q = K α A A - 4 where K α is the kinetic energy of the α -particle.

Given the reaction X Z A → Y Z - 2 A - 4 + He 2 4 + Q To prove that Q = K α A A - 4 The kinetic energy of α particle is given by K α = mass no . after decay mass no . before decay × Q = A - 4 A Q The Q value is given by Q = M α - M y M y × K α = 4 - A - 4 A - 4 × Q = A A - 4 A - 4 A Q = Q hence proved.

α - decay is a process in which the nucleus of a radioactive element emits an alpha particle. α - particle is a doubly ionised helium atom H 2 4 e + + so by emitting an alpha particle the mass number of the daughter nucleus is reduced by 4 and the atomic number is reduced by 2 . A general alpha decay process can be represented by the following equation.: X Z A → Y Z - 2 A - 4 + He 2 4 This spontaneous emission of the α particle is possible only when the total mass of decay products is less than that of the initial nucleus.

In the reaction X Z A → Y Z - 2 A - 4 + H 2 4 e + Q of the nucleus X at rest. Taking the ratio of the mass of the α - particle M u and the mass of Y nucleus as M U M Y = 4 A - 4 show that the Q value of the reaction is given by Q = K α A A - 4 where K α is the kinetic energy of the α -particle.

Given the reaction X Z A → Y Z - 2 A - 4 + He 2 4 + Q To prove that Q = K α A A - 4 The kinetic energy of α particle is given by K α = mass no . after decay mass no . before decay × Q = A - 4 A Q The Q value is given by Q = M α - M y M y × K α = 4 - A - 4 A - 4 × Q = A A - 4 A - 4 A Q = Q hence proved.

Write the equation of proton-proton cycle in the sun.

The proton-proton cycle is a chain of thermonuclear reactions that is the chief source of the energy radiated by the Sun and other cool main-sequence star s.It has the following steps: H 1 1 + H 1 1 ⇒ H 1 2 + e + + ν + 0 . 42 MeV .......................................................(1) e + + e - ⇒ γ + γ + 1 . 02 MeV .....................................................................(2) H 1 2 + H 1 1 ⇒ He 2 3 + γ + 5 . 49 keV ...................................................................(3) He 2 3 + He 2 3 ⇒ He 2 4 + H 1 1 + H 1 1 + 12 . 86 MeV .............................................(4) For the fourth reaction to occur, the first three reactions must occur twice, in which case two light He nuclei unite to form an ordinary helium nucleus. If we consider 2 1 + 2 2 + 2 3 + 4 ,The net effect is: 4 H 1 1 + 2 e - ⇒ He 2 4 + 2 ν + 6 γ + 26 . 7 MeV