Solved Numerical Problem No. 21.6 ( Unit-21:Nuclear Physics); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

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)

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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