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प्रश्न
Potassium-40 can decay in three modes. It can decay by β−-emission, B*-emission of electron capture. (a) Write the equations showing the end products. (b) Find the Q-values in each of the three cases. Atomic masses of `""_18^40Ar` , `""_19^40K` and `""_20^40Ca` are 39.9624 u, 39.9640 u and 39.9626 u respectively.
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
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उत्तर
(a) Decay of potassium-40 by β−emission is given by
`""_19"K"^40` → `""_20"Ca"^40` + `β^-` + `bar"v"`
Decay of potassium-40 by β+ emission is given by
`""_19"K"^40 → ""_18"Ar"^40 + β^+ + "v"`
Decay of potassium-40 by electron capture is given by
`""_19"K"^40 + e^(-) → ""_18"Ar"^40 + "v"`
(b)
Qvalue in the β− decay is given by
Qvalue = [m(19K40) − m(20Ca40)]c2
= [39.9640 u − 39.9626 u]c2
= 0.0014 `xx` 931 MeV
= 1.3034 MeV
Qvalue in the β+ decay is given by
Qvalue = [m(19K40) − m(20Ar40) − 2me]c2
= [39.9640 u − 39.9624 u − 0.0021944 u]c2
= (39.9640 − 39.9624) 931 MeV − 1022 keV
= 1489.96 keV − 1022 keV
= 0.4679 MeV
Qvalue in the electron capture is given by
Qvalue = [ m(19K40) − m(20Ar40)]c2
= (39.9640 − 39.9624)uc2
= 1.4890 = 1.49 MeV
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