Answer:
Explanation:
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In this case, by considering helium gas as an ideal one, we can use the following equation:
Whereas P should go in atmospheres and T in Kelvins; thus we proceed as follows:
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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.
