1.

Find the Q-value of the emitted alpha-particle in the alpha-decay of ._(88)^(226)Ra.m_(alpha)=4.00260u m(._(88)^(226)RA)=226.02540u m(._(86)^(222)Rn)=222.01750u

Answer»

8.93 MeV
4.93 MeV
12.3 MeV
None of these

Solution :`DELTAM=(226.02540)-(222.01750+4.00260)`
`Deltam=.0053am U`
`Q=Deltamxx931.25`
`Q=4.93MeV`


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