HW5-2-3 - 5-3 2 The radioisotope 224Ra ecaysby a emission...

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5-3 2. The radioisotope 224Ra decays by a emission primarily to the ground state of 220Rn (94% probability) and to the first excited state 0.241 MeV above the ground state (5.5% probability). What are the energies of the two associated a particles? . Solution: For the daughter left in an excited state, the Q- 224Ra value of Eq. (5.7) is modified to increase the rest ~ 5~ mass of M(2~gRn) by an amount E* /c2, namely .at . ~ 94.5~ Qal = {Mr~:Ra) -[M(~He) +M(2~gRn) + E* JC2]} C2, ~(..Ql~ ~J_--~~ With the masses in Ap. B we find Qal = {224.0202020 -[4.00260325 + 220.0113841]} (u) x 931.5 (MeV lu) -0.241 MeV = 5.548 MeV. The kinetic energy of Ql is given by Eq. (5.11), namely [ M(220Rn) ] [ 220 ] Eal = Qal M(2~~Rn)8~ M(~He) ~ 5.548 m = 5.449 MeV. In a similar manner for alpha decayto the ground state (E* = 0), we have
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