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Obtain the maximum kinetic energy of bet...

Obtain the maximum kinetic energy of `beta`-particles, and the radiation frequencies of `gamma` decays in the decay scheme shown in Fig. `14.6`. You are given that `m(.^(198)Au)=197.968233 u, m(.^(198)Hg)=197.966760 u`

Text Solution

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`gamma`-rays are elctro magnetic radiations having energy `E=h upsilonrArrupsilon=(E)/(h)` where `h=`plank's constant`=6.625xx10^(-34)J.S`
1) Frequencies of `gamma_(1),gamma_(2)` and `gamma_(3)` are calculated as follows
`gamma_(1)=(DeltaE)/(h)=((1.088-0)MeV)/(6.625xx10^(-34)J.s)`
`=(1.088xx10^(6)xx1.6xx10^(-19))/(6.625xx10^(-34))`
`0.2627xx10^(21)=2.627xx10^(20)Hz`
`gamma_(2)=(DeltaE)/(h)=((0.412-0)MeV)/(6.625xx10^(-34)Js)`
`=(0.412xx1.6xx10^(-19)xx10^(6))/(6.625xx10^(-34))`
`=0.0995xx10^(21)=9.95xx10^(19)Hz`
`gamma_(3)=(DeltaE)/(h)=((1.088-0.412)xx10^(6)xx1.6xx10^(-19))/(6.625xx10^(-34))`
`=0.1632xx10^(21)=1.62xx10^(20)Hz`
2) Now maximum `K.E` of `beta_(1)^(-)=[M(.^(198)._(79)Au)M(.^(198)._(80)Hg)-(1.088)/(931.5)]c^(2)`
`(because 1 amu = 931.5 MeV rArr1 MeV=(1)/(931.5)U)`
`=[197.968233-197.966760-0.00168]931.5MeV`
`=0.000305xx931.5=0.284 MeV`
Maximum `K.E` of `beta_(2)^(-)=`
`[M(.^(198)._(79)Au)-M(.^(198)._(80)Hg)-(0.412)/(931.5)]c^(2)`
`=[197.968233-197.966760-0.000442]931.5`
`=0.001031xx931.5=0.9603 MeV`
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