Nuclear Physics And Radioactivity Question 328
Question: A radioactive element $ _{90}X^{238} $ decay into $ _{83}Y^{222} $ . The number of $ \beta - $ particles emitted are [BHU 1997; JIPMER 2001, 02]
Options:
A) 4
B) 6
C) 2
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
Number of $ \alpha - $ particles emitted $ =\frac{238-222}{4}=4 $ This decreases atomic number to $ 90-4\times 2=82 $ Since atomic number of $ _{83}Y^{222} $ is 83, this is possible if one $ \beta - $ particle is emitted.
BETA
