A) \[_{40}^{91}Zr\]
B) \[_{\,\,36}^{101}Kr\]
C) \[_{\,\,36}^{103}Kr\]
D) \[_{\,\,56}^{144}Ba\]
Correct Answer: D
Solution :
[d] \[\bigcup{_{92}^{235}}+_{0}^{1}n\to Kr_{36}^{89}+3n_{0}^{1}+X_{Z}^{A}\] |
\[92+0=36+Z\] |
\[\Rightarrow Z=56\] |
\[235+1=89+3+A\] |
\[\Rightarrow A=144\] |
So, \[_{\,\,56}^{144}Ba\] is generated. |
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