1.

Find the Q-value and the kinetic energy of the emitted alpha - particle in the alpha - decay of ._(88)^(226)Ra Given m (._(88)^(226)Ra)=226.02540 u, m (._(86)^(222)Rn)=222.01750 u, m (._(86)^(222)Rn)=220.01137 u, m (._(84)^(216)Po) = 216.00189 u.

Answer»

Solution :`._(88)Ra^(226)rarr ._(85)RN^(222)+._(2)He^(4)`
Q VALUE
`[m(._(88)Ra^(226))-m(._(86)Rn^(222))-m_(ALPHA)]xx931.5 MeV`
`= [226.02540-222.01750-4.00260]xx931.5 MeV`
`Q = 0.0053xx931.5 MeV = 4.94 MeV`
K.E of a particle `= ((A-4)Q)/(A)=(226-4)/(226)xx4.94 = 4.85 MeV`


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