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)=226.025

Question

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)=226.025

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`.

Monjur Alahi

7 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

`._(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`

7 views Answered 2023-02-05 15:29