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

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Unformatted text preview: 5-3 2. The radioisotope 224Ra ecaysby a emission primarily to the ground state of d 220Rn (94% probability) and to the first excited state 0.241 MeV above the ground state (5.5% probability). What are the energiesof the two associated a particles? . Solution: For the daughter left in an excited state, the Qvalueoff Mq. (5.7) is y an amountincreasenhe rest mass o E (2~gRn) b modified to E* /c2, t amely Qal = {Mr~:Ra) -[M(~He) ~ 224Ra .at .5~ ~ 94.5~ ~(..Ql~ +M(2~gRn) + E* JC2]} C2, ~J_--~~ With the massesin 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 Eal = Qal [M(2~~Rn)8~ M(~He) M(220Rn) ~ 5.548 220 = 5.449 MeV. m In a similar manner for alpha decayto the ground state (E* = 0), we have Qa2 = {M(2~~Ra) -[M(~He) + M(2~gRn)]} c2 = {224.0202020 [4.00260325 + 220.0113841]} 31.5 (MeV /u) 9 = 5.789 MeV. The kinetic energy of this alpha particle is [ , [I M(220Rn) --- Ea2 = Qa2 M(2~gRn)8~ M(~He) 3. The radionuclide 41 r decays by {3- emission to an excited level of 41 that A K is 1.293 MeV above the ground state. What is the maximum kinetic energy of the emitted .8- particle? Solution: First find the Qf3- for this decay from Eq. (5.15) with the atomic massesin Ap. B. The result is Qf3- = {M(1~Ar) -[M(1AK) + E* jC2]} C2 x 931.5 41Ar - " ~(.112~ = [40.9645008 -40.96182597] -1.293 MeV = 1.199 MeV. _J___~L The maximum kinetic energy the beta particle can have equals this Q-value, i.e., (E{3)max 1.199 MeV. = July 24, 2002 "'( ...
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