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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 (^{198}Au) = 197.968233 u

m (^{198}Hg) =197.966760 u

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#### Solution

It can be observed from the given γ-decay diagram that γ_{1} decays from the 1.088 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ_{1}-decay is given as:

E_{1} = 1.088 − 0 = 1.088 MeV

hν_{1}= 1.088 × 1.6 × 10^{−19} × 10^{6} J

Where,

h = Planck’s constant = 6.6 × 10^{−34} Js

ν_{1 }= Frequency of radiation radiated by γ_{1}-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 γ_{2} decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ_{2}-decay is given as:

E_{2} = 0.412 − 0 = 0.412 MeV

hν_{2}= 0.412 × 1.6 × 10^{−19} × 10^{6} J

Where,

ν_{2 }= Frequency of radiation radiated by γ_{2}-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 γ_{3} decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

Hence, the energy corresponding to γ_{3}-decay is given as:

E_{3} = 1.088 − 0.412 = 0.676 MeV

hν_{3}= 0.676 × 10^{−19} × 10^{6} J

Where,

ν_{3 }= Frequency of radiation radiated by γ_{3}-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/c^{2}

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"`

β_{1 }decays from the 1.3720995 MeV level to the 1.088 MeV level

∴ Maximum kinetic energy of the β_{1 }particle = 1.3720995 − 1.088

= 0.2840995 MeV

β_{2 }decays from the 1.3720995 MeV level to the 0.412 MeV level

∴ Maximum kinetic energy of the β_{2 }particle = 1.3720995 − 0.412

= 0.9600995 MeV

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