Find the Q-value and the kinetic energy of the emitted `alpha` - particle in the `alpha` - decay of
`._(86)^(220)Rn`.
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`.

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1 Answers

Proceeding as above, in case of `._(86)Rn^(220)`
Q = 6.41 MeV
K.E of a particle
`= ((A-4)Q)/(A) = ((220-4))/(220)xx6.41 = 6.29 MeV`

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