(The radioisotope 224Ra decays by alpha emission via two paths to the ground state

a) ²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q

b) the Q-value of this reaction is 5.789 MeV

c) the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev

Explanation:

a)

The decay equation for the alpha decay is expressed as;

²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q

b)

Calculate the Q-value (in MeV) of this reaction.

Q = Mparent - Mdaughter -Mg

Q = MRa - MRn -Mg

= 224.020202 - 220.011384 - 4.00260305

= 0.00621495 amu

= 5.789 MeV

therefore the Q-value of this reaction is 5.789 MeV

c)

Energy of alpha particle is expressed as;

E∝ = MQ / ( m + M)

now this is the maximum energy available for the daughter, ²²⁰Rn going to the ground state;

The energy of the alpha particle gives;

E∝  = 220(5.789) / ( 4 + 220) = 5.69 MeV

as given in the question,The other less frequent alpha occurring 5.5% of the time leaves the daughter nucleus in an excited state of 0.241 MeV above the ground state.

Therefore the energy of this alpha is

E∝ = 5.69 - 0.241 = 5.449 Mev

Therefore the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev

d)

Sketch of the nuclear decay scheme have been uploaded along side this answer.