Question

(The radioisotope 224Ra decays by alpha emission via two paths to the ground state of its daughter 94% probability of alpha decay directly to the daughter ground state;(20 pts, 5 pts each)5.5% probability of alpha decay to an excited state before emitting a 0.241 MeV photon in an isomeric transition to reach the ground state of the daughter.a.Write the reaction equation for the alpha decay of 224Ra. b.Calculate the Q-value (in MeV) of this reaction. c.What are the energies (in MeV) of the two associated alpha particles? d.Sketch the decay scheme for the decay of 224Ra with the associated probabilities, energies, and emissions for the a

Answer 1

Answer:

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.