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The nucleus ""(10)^(23) Ne decays by be...

The nucleus `""_(10)^(23)` Ne decays by `beta–` emission. Write down the -decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
`m(""_(10)^(23)Ne)= 22.994466u`
`m(""_(11)^(23)Na)=22.089770 u.`

Text Solution

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The equation representing `beta`-decay of `._(10)^(23)Ne` is
`._(10)^(23)Nerarr._(11)^(23)Na+beta+overset(-)v` where `Q` is the kinetic shared by `._(10)^(23)Ne` and `._(11)^(23)Na`. Ignoring the rest mass of antineutrino `(overset(-)v)` and electron.
Mass defect `Delta m`
`=m(._(10)^(23)Ne)-m(._(11)^(23)Ne)(._(11)^(23)Na)-m(beta^(-))`
`22.994466-22.989770=0.004696u`
`:. Q = 0.004696xx931MeV=4.372 MeV`
`:.` Max. `K.E` of `beta^(-)=4.372 MeV` when energy carried by is zero.
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