A nucleus with mass number 220 initially at rest emits an alpha particle. If the Q value of the reaction is 5.5 MeV, the kinetic energy of the alpha particle is (a) 5.6 MeV (b) 4.4 MeV (c) 6.5 MeV (d) 5.4 MeV

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 M_x. The mass number of alpha particle is 4 and is denoted by M_α. Hence, we have

M_x = M – M_α

On substituting we get,

M_x = 220 – 4 = 216

Here, Q value of the reaction is total energy and is given by

Q = E_α + E_x

where,

E_α is the energy of an alpha particle and E_x is the energy of daughter nuclei

As per the law of conservation of momentum

p_α = p_x

where,

p_α is the momentum of alpha particle adn p_x 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

m_αE_α = m_xE_x

E_α = (m_x/m_α)Q

E_α = 216/220 x 5.5

We get,

E_α = 5.4 MeV

Therefore, the correct option is (d)