Contoh Soal Dan Penyelesain Ke2

Problem Set #2

37 1. (a) Calculate the Q value for K orbital-electron capture by the 18  Ar  nucleus,

neglecting the electron binding energy. (b) Repeat (a), including the binding energy, 3.20 keV, of the K-shell electron in argon. (c) What becomes of the energy released as a result of this reaction? 224

220

2. The radioisotope Ra decays by α emission primarily to the ground state of Rn (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 α particles? 41

41

−

3. The radionuclide Ar decays by  β  emission to an excited level of K that is 1.293 MeV above the ground state. What is the maximum maximum kinetic energy of the emitted  β   particle? 64

−

−

4. The radioisotope Cu decays by three different mechanisms:  β  decay (39.0%), +

electron capture (EC) (43.1%) and  β  decay (17.4 %). The Q value for the  β  +

−

decay is

578.7 keV. The Q value for the  β  decay is 653.1 keV. In addition, there is a gamma emission (0.5% probability) at 1.345 MeV. Sketch the energy level diagram for the decay scheme.

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1) (# points) a). The decay equation is: 37 18

0

 Ar + 1 e −

−

37 →17

Cl + υ 

 Neglecting the binding energy of the electron: QEC = Δ parent - Δdaughter   = ΔAr – Δ Ni = -30.951-(-31.765) = 0.814 MeV

(3.25)

 b). Including now the electron binding energy 3.2 keV = .0032 MeV gives: QEC = Δ parent - Δdaughter -EB = 0.814-0.0032 = 0.8108 MeV

B

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c). The neutrino will take the entire energy, the Q energy, from the reaction since it is practically mass less.

2. ( # points) The decay equation for the alpha decay is: 224 88

220

 Ra → α + 86 Rn + Q

Calculating the Q value of the reaction, written with masses in amu: Q = M parent – Mdaughter  -Mα = MRa – MRn -Mα = 224.020202-220.011384-4.00260305 = 0.00621495 amu = 5.789 MeV

(3.12)

The energy of the alpha particle is:

 E α 

=

 MQ m +  M 

This is the maximum energy available for the daughter, Calculating the energy of the alpha particle gives:

 E α 

=

220(5.789) 4 + 220

=

220

Rn going to the ground state.

5.69 MeV 

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 α  = 5.69 − 0.241 = 5.449 MeV  The nuclear decay scheme that shows this is: 224

Ra

α 5.69 MeV (94%)

5.69 MeV α 5.449 MeV (5.5%) 0.241 MeV

220

Rn

γ 0 MeV

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3. ( # points) The decay equation is: 41 18

41

0

 Ar →19 K + 1 β  −

−

+

υ  + Q

Calculating the Q energy, using masses: Q = M parent – Mdaughter = MAr  – MK = 40.964500-40.961825 = 0.002675 amu = 2.492 MeV The beta emission here leaves the daughter 41K atom in an excited state of 1.293 MeV. Therefore the remaining energy will go to the beta minus and the antineutrino, hence 1.2 MeV. The maximum energy that the emitted beta minus will have is then the total remaining energy with the antineutrino receiving nothing, therefore the maximum kinetic energy is 1.2 MeV.

4. (# points) The decay scheme is:

64

Cu

1.6731 MeV EC (0.5%)

2

2mc

1.345 MeV

64

Cu

0.6531 MeV

+

β (17.4%) 64

 Ni

EC (42.6%)

64

γ (0.5%)

Cu

0.5787 MeV

-

0 MeV

β (39%) 64

Zn

0 MeV