Unformatted text preview: 53 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 Qvalue, i.e., (E{3)max 1.199 MeV. =
July 24, 2002 "'( ...
View
Full
Document
This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell.
 Spring '06
 CADY

Click to edit the document details