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.(92)U^(238) is a natural alpha-emitter....

`._(92)U^(238)` is a natural `alpha`-emitter. After `alpha`-emission the residual nucleus `U_(X1)` in turn emits a `beta`-particle to produce another nucleus `U_(X2)`. Find out the atomic number and mass number of `U_(X1)` and `U_(X2)`.

Text Solution

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Parent element = `._(92)U^(238)`
After `alpha`-emission, the daugther nucleus formed is `U_(X1)`.
The reaction is:
`._(92)U^(238) rarr ._(Z)U_(XI)^(A) + He^(4)`
Equations atomic mass on both sides, we get
`238 = A + 4`
`:. A = 234`
Equation atomic mass on both sides, we get
`92 = Z + 2`
`:. Z = 90`
The product is `._(90)U^(234)_(XI)`. This product atom emits `beta`-particle
`._(90)U^(234) rarr ._(Z)U_(X2)^(A) + ._(-1)e^(0)`
Solving for atomic mass, we get `A = 234`
Solving for atomic number, we get `Z = 91`
Therefore, the product atom is `._(91)U_(X2)^(234)`.
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