Nuclear Physics And Radioactivity Question 280
Three $ \alpha $-particles and one $ \beta $-particle decaying takes place in series from an isotope $ _{88}Ra^{238} $ . Finally the isotope obtained will be [CPMT 1989; DCE 2000]
Options:
A) $ _{84}X^{220} $
B) $ _{86}X^{222} $
C) $ _{83}X^{224} $
D) $ _{83}Bi^{215} $
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Answer:
Correct Answer: C
Solution:
By using $ {n_{\alpha }}=\frac{A-A’}{4} $ and $ {n_{\beta }}=2{n_{\alpha }}-Z+Z’ $
Þ $ A’=A-4{n_{\alpha }}=236-4\times 3=224 $ and $ Z’=({n_{\beta }}-2{n_{\alpha }}+Z)=(1-2\times 3+88)=83 $
BETA
