Question

The particle $X$ in the following nuclear reaction is ${}_7^{13}\text{N}$ $⟶$ ${}_6^{13}\text{C}+$ ${}_1^0\text{e}$ + $X$

• A. $P$
• B. $v$
• C. $e^{-}$
• D. $\alpha$

Answer 1

The correct option is B $v$

The given reaction : ${}_7^{13}N{\to }_6^{13}C{+}_1^{0}e{+}_Z^{A}X$

Conservation of mass number : $13=13+0+A$ $⟹A=0$

Conservation of atomic number : $7=6+1+Z$ $⟹Z=0$

Thus, the particle $X$ is neutrino i.e. $\nu$