Question

what should go into the empty box in this equation?
103 44 ru→□+103 45 rh
4 2 α
0 -1 β
an alpha particle
a beta particle

Explanation:

Step1: Check mass number balance

The mass number of \(^{103}_{44}Ru\) is \(103\), and the mass number of \(^{103}_{45}Rh\) is \(103\). For mass - number balance (\(A_{reactant}=A_{product1} + A_{product2}\)), if we assume the particle is \(^{A}_{Z}X\), then \(103=A + 103\), so \(A = 0\).

Step2: Check atomic number balance

The atomic number of \(^{103}_{44}Ru\) is \(44\), and the atomic number of \(^{103}_{45}Rh\) is \(45\). For atomic - number balance (\(Z_{reactant}=Z_{product1}+Z_{product2}\)), \(44=Z + 45\), so \(Z=- 1\).

A beta particle (\(^{\ 0}_{-1}\beta\)) has \(A = 0\) and \(Z=-1\), while an alpha particle (\(^{4}_{2}\alpha\)) has \(A = 4\) and \(Z = 2\).

Answer:

A beta particle.