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^{28}Th emits an alpha particle to reduce to ^{224}Ra. 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 ^{228}Th is 228.028726 u, that of ^{224}Ra is 224.020196 u and that of `""_2^4H` is 4.00260 u.

(Use Mass of proton m_{p} = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron m_{n} = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c^{2},1 u = 931 MeV/c^{2}.)

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#### Solution

Given:-

Atomic mass of ^{228}Th, *m*(^{228}Th) = 228.028726 u

Atomic mass of ^{224}Ra, *m*(^{224}Ra) = 224.020196 u

Atomic mass of `""_2^4H, m(""_2^4H) = 4.00260 "u"`

Mass of ^{224}Ra = 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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