If `\ _{92}^{233}U` decays twice by ∝-emission. what is the resulting isotope?
(Ans : `\ _{88}^{225}Rn` )
Given Data:
Element = `\ _{92}^{233}U`
∝-emission (twice)
∝-emission (twice)
To Find:
Resulting Isotopes = ?
Solution:
The general reaction for ∝-emission is
`\ _{Z}^{A}X` → `\ _{Z-2}^{A-4}X` + `\ _{2}^{4}∝` + Energy
According to the given conditions, the reaction will be as follows (After 1st ∝-emission)
`\ _{92}^{233}U` → `\ _{90}^{229}Th` + `\ _{2}^{4}∝` + Energy
After 2nd ∝-emission
`\ _{90}^{229}Th` → `\ _{88}^{225}Rn` + `\ _{2}^{4}∝` + Energy
Thus the remaining isotopes are
`\ _{88}^{225}Rn` ----------------Ans
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