In the nuclear reaction, \({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+\beta+X, X\) stands for

In the nuclear reaction, \({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+\beta+X, X\) stands for
  1. a neutron
  2. an electron
  3. a neutrino
  4. an anti-neutrino

Solution

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.

Asked in: AP EAMCET 2020 (17 Sep Shift 1)

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