Let M be the nucleus with mass number and X be the daughter nuclei produced.
Hence, we have
M → X +α
Let the mass of daughter nuclei be Mx. The mass number of alpha particle is 4 and is denoted by Mα. Hence, we have
Mx = M – Mα
On substituting we get,
Mx = 220 – 4 = 216
Here, Q value of the reaction is total energy and is given by
Q = Eα + Ex
where,
Eα is the energy of an alpha particle and Ex is the energy of daughter nuclei
As per the law of conservation of momentum
pα = px
where,
pα is the momentum of alpha particle adn px is the momentum of daughter nuclei
We know,
The relationship of energy and momentum is given by
E = p2/2m
Hence, from the law of conservation of momentum, we have
mαEα = mxEx
Eα = (mx/mα)Q
Eα = 216/220 x 5.5
We get,
Eα = 5.4 MeV
Therefore, the correct option is (d)