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.
(a)[tex]^{224}Ra_{_{88}}=^4\alpha_{_2}+^{220}Rn_{_{86}}+Q[/tex]
(b) the Q-value of the reaction is 5.789 MeV
(c) The alpha particles have energies of 5.69 MeV and 5.449 MeV
(a) The alpha decay of the radioisotope 224Ra is given by:
[tex]^{224}Ra_{_{88}}=^4\alpha_{_2}+^{220}Rn_{_{86}}+Q[/tex]
(b) The Q-value of a reaction is given by:
Q = {mas of reactants - mass of products}c²
[tex]Q = [M_{224Ra} - M_{\alpha} - M_{220Rn}]c^2[/tex]
Q = (224.020202 - 220.011384 - 4.00260305) amu.c²
Q = 0.00621495 × 931.5 MeV ( since 1 amu.c² = 931.5 MeV)
Q = 5.789 MeV
(c) The energy of the alpha particle is expressed as;
E(α) = MQ / ( m + M)
it is the maximum energy available for the daughter nucleus, ²²⁰Rn going to the ground state.
If the daughter nucleus is at rest then all the energy is carried by the alpha particle.
The energy of the alpha particle is:
E(α) = 220(5.789) / ( 4 + 220) = 5.69 MeV
Now, for the alpha particle occurring 5.5% of the time and leaving the daughter nucleus in an excited state of 0.241 MeV :
The energy of this alpha particle is
E(α) = 5.69 - 0.241 = 5.449 Mev
Learn more about alpha decay:
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