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




************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.