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Question
Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γdecays in the decay scheme shown in Fig. 13.6. You are given that
m (198Au) = 197.968233 u
m (198Hg) =197.966760 u
Solution
It can be observed from the given γ-decay diagram that γ1 decays from the 1.088 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ1-decay is given as:
E1 = 1.088 − 0 = 1.088 MeV
hν1= 1.088 × 1.6 × 10−19 × 106 J
Where,
h = Planck’s constant = 6.6 × 10−34 Js
ν1 = Frequency of radiation radiated by γ1-decay
`therefore "v"_1 = "E"_1/"h"`
`= (1.088 xx 1.6 xx 10^(-19) xx 10^6)/(6.6 xx 10^(-34)) = 2.637 xx 10^20 "Hz"`
It can be observed from the given γ-decay diagram that γ2 decays from the 0.412 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ2-decay is given as:
E2 = 0.412 − 0 = 0.412 MeV
hν2= 0.412 × 1.6 × 10−19 × 106 J
Where,
ν2 = Frequency of radiation radiated by γ2-decay
`therefore "v"_2 = "E"_2/"h"`
`= (0.412 xx 1.6 xx 10^(-19) xx 10^6)/(6.6 xx 10^(-34)) = 9.988 xx 10^(19) "Hz"`
It can be observed from the given γ-decay diagram that γ3 decays from the 1.088 MeV energy level to the 0.412 MeV energy level.
Hence, the energy corresponding to γ3-decay is given as:
E3 = 1.088 − 0.412 = 0.676 MeV
hν3= 0.676 × 10−19 × 106 J
Where,
ν3 = Frequency of radiation radiated by γ3-decay
`therefore "v"_3 = "E"_3/"h"`
`= (0.676 xx 1.6 xx 10^(-19) xx 10^6)/(6.6 xx 10^(-34)) = 1.639 xx 10^20 "Hz"`
Mass of `"m"(""_78^198"Au")`= 197.968233 u
Mass of `"m"(""_80^198 "Hg")`= 197.966760 u
1 u = 931.5 MeV/c2
Energy of the highest level is given as:
E = `["m"(""_78^198 "Au") - "m"(""_80^190 "Hg")]`
`= 197.968233 - 197.966760 = 0.001473 " u"`
`= 0.001473 xx 931.5 = 1.3720955 " MeV"`
β1 decays from the 1.3720995 MeV level to the 1.088 MeV level
∴ Maximum kinetic energy of the β1 particle = 1.3720995 − 1.088
= 0.2840995 MeV
β2 decays from the 1.3720995 MeV level to the 0.412 MeV level
∴ Maximum kinetic energy of the β2 particle = 1.3720995 − 0.412
= 0.9600995 MeV
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