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Question
28Th emits an alpha particle to reduce to 224Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:
`""^228"Th" → ""^224"Ra"^(∗) + alpha`
`""^224"Ra"^(∗) → ""^224"Ra" + γ (217 "keV")`.
Atomic mass of 228Th is 228.028726 u, that of 224Ra is 224.020196 u and that of `""_2^4H` is 4.00260 u.
(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.)
Solution
Given:-
Atomic mass of 228Th, m(228Th) = 228.028726 u
Atomic mass of 224Ra, m(224Ra) = 224.020196 u
Atomic mass of `""_2^4H, m(""_2^4H) = 4.00260 "u"`
Mass of 224Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV
Kinetic energy of alpha particle, K = `[m(""^228"Th") - [m(""^224"Ra") + m(""_2^4"H")]]c^2`
= (228.028726 × 931) − [(208563.0195 + 4.00260 × 931]
= 5.30383 MeV = 5.304 MeV
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