`^228Th`emits an alpha particle to reduce to `^224Ra`.Calculate the kinetic energy of the alpha particle emitted in the following decay:
`^228Th rarr ``224Ra**+ alpha`
``224Ra** rarr `224Ra+ gamma(217 keV).`
Atomic mass of ``228Th` is 228.028726 u,that of ``224Ra`is 224.020196 u and that of ``_2^4He` is 4.00260 u.

Answer

C, D

Explanation

` ^(228)Th rarr ^(224) Ra^(**) + alpha `
` ^(224) Ra^(**) rarr ^(224) Ra + v(217Kev)`
` [Mass
` ^(228) Th rarr 228.028736u`
` ^(224) Ra rarr 224.020196 u`
`alpha = _(2) ^(4) He rarr 4.002650 u]`
Now mass of
` ^(224)Ra = 224.020196 xx 931 + 0.217 MeV`
` = 208563.0196 MeV`
` KE of 'alpha' = E (^(228)Th) - E (^(224)Ra^(**)+ alpha)`
`= 228.028726 xx 931 - `
` [208563.0195 + 4.00260 xx 931] `
` = 5.30383 MeV = 5.304 MeV ` .