Question

ln the nuclear reaction $_7^{14} N + X \rightarrow _6^{14} C + _1^1 H,$ the $X$ will be

• A. $_{-1}^{0} e$
• B. $_1^1 H$
• C. $_1^2 H$
• D. $_0^1 n$

Answer 1

Answer: (D)

$_0^1 n$

Answer 2

In the reaction, we have to conserve the mass number and the atomic number. thus, we have
$_7^{14} N +\, _b^a X \rightarrow\, _6^{14} C + \,_1^1 H$
So, $14 + a = 14 + 1$
Hence, $a = 1$ and $7 + b = 6 + 1$
Hence, $b = 0$
So, we have $_b^a X =\, _0^1 X = \,_0^1 n$