Question

what is the missing particle?
{}_{6}^{14}c \to {}_{7}^{14}n + \square ?

Explanation:

Step1: Calculate mass number

Mass number of \(_{6}^{14}\text{C}\) is \(14\), mass number of \(_{7}^{14}\text{N}\) is \(14\). Let mass number of missing particle be \(A\). Then \(14 = 14+A\), so \(A = 0\).

Step2: Calculate atomic number

Atomic number of \(_{6}^{14}\text{C}\) is \(6\), atomic number of \(_{7}^{14}\text{N}\) is \(7\). Let atomic number of missing particle be \(Z\). Then \(6 = 7+Z\), so \(Z=- 1\).

Step3: Identify particle

A particle with \(A = 0\) and \(Z=-1\) is \(\beta\) particle (\(_{-1}^{0}\text{e}\)).

Answer:

\(_{-1}^{0}\text{e}\)