`Th^228`emits an alpha particle to reduce to `Ra^224`.Calculate the kinetic energy of the alpha particle emitted in the following decay:
`Th^228 rarr Ra^**+ alpha`
`(Ra^224)^** rarr Ra^224+ gamma(217 keV)`
Atomic mass of `Th^228` is `228.028726 u`,that of `Ra^224`is `224.020196 u` and that of `He_2^4` 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 ` .