A) 140
B) 144
C) 142
D) 146
Correct Answer: B
Solution :
\[_{92}{{M}^{238}}\xrightarrow{{}}{{\,}_{y}}{{N}^{x}}+2{{\,}_{2}}H{{e}^{4}}\] \[_{y}{{N}^{x}}\xrightarrow{{}}{{\,}_{B}}{{L}^{A}}+2{{\beta }^{+}}\] \[_{y}{{N}^{x}}{{=}_{(92-2\times 2)}}{{N}^{(238-4\times 2)}}={{\,}_{88}}{{N}^{230}}\] \[_{88}{{N}^{230}}\xrightarrow{2{{\beta }^{+}}}{{\,}_{(88-2)}}{{L}^{(230)}}={{\,}_{86}}{{L}^{230}}\] Total no of neutrons in \[_{90}{{L}^{330}}\] \[230-86=144\]You need to login to perform this action.
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