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Find the Q value and the kinetic energy ...

Find the Q value and the kinetic energy of emitted `alpha` particle in the `alpha` decay of
(a) `._88Ra^(226)` (b) `._86Rn^(220)`. Given `m(._88Ra^(226))=226.02540u, m(._86Rn^(222))=222.01750u`
(b) `m(._86Rn^(220))=220.01137u , m(._84Po^(216))=216.00189u` and `m._(alpha)=4.00260u.`

Text Solution

Verified by Experts

(a) `._88Ra^(226) to ._86Rn^(222)+._2He^4`
Q value =`[m(._88Ra^(226))-m(._86Rn^(222))-m_(alpha)]xx931.5MeV`
`=[226.02540-222.01750-4.00260]xx931.5MeV`
`Q=0.0053xx931.5MeV=4.94MeV`
K.E. of `alpha` particle `=((A-4)Q)/A=(226-4)/226xx4.94=4.85MeV`
(b) Proceeding as above, in case of `._86Rn^(220)`
`Q=6.41MeV`
K.E. of `alpha` particle =`((A-4)Q)/A=((220-4))/220xx6.41=6.29 MeV`
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