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 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)

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