Question

How much energy is released in the following reaction : ⁷Li + p → α + α.
Atomic mass of ⁷Li = 7.0160 u and that of ⁴He = 4.0026 u.

(Use Mass of proton m_p = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron m_n = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c²,1 u = 931 MeV/c².)

Answer 1

Given:-
Mass of ⁷Li = 7.0160 u
Mass of ⁴He = 4.0026 u.

Reaction :-

`L_i^7 + p → alpha + alpha + E` ,

Energy release (E) is given by

`E = [m(""^7L_i) + (m_p) - 2 xx m ("^4H_e)] c^2`

`= [(7.0160  "u" + 1.007276  "u") - 2(4.0026  "u")]c^2`

`= (8.023273  "u" - 8.0052  "u") c^2`

`= 0.018076 xx 931  "MeV"`

`= 16.83  "MeV"`