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In the nuclear reaction, \({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+\beta+X, X\) stands for
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The correct answer is:
a neutrino
Nuclear reaction is given as
\({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+\beta+X\) ...(i)
When positron ( \(\beta\) ) is emitted from a nuclei, then atomic number decreases by one unit, while the mass number remains the same.
Hence, Eq. (i) is written as
\({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+{ }_1^0 \beta+v \text { (neutrino) }\)
Hence, \(X\) is a neutrino v.
\({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+\beta+X\) ...(i)
When positron ( \(\beta\) ) is emitted from a nuclei, then atomic number decreases by one unit, while the mass number remains the same.
Hence, Eq. (i) is written as
\({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+{ }_1^0 \beta+v \text { (neutrino) }\)
Hence, \(X\) is a neutrino v.
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