Question

what should go into the empty box in this equation? *
293 116 lv→□+289 114 fi
4 2 α
0 -1 β
an alpha particle
a beta particle
there is not enough information to answer this question

Explanation:

Step1: Check mass number balance

The mass number of \(_{116}^{293}Lv\) is \(293\), and the mass number of \(_{114}^{289}Fl\) is \(289\). Let the mass number of the emitted particle be \(x\). Then \(293=x + 289\), so \(x=293 - 289=4\).

Step2: Check atomic number balance

The atomic number of \(_{116}^{293}Lv\) is \(116\), and the atomic number of \(_{114}^{289}Fl\) is \(114\). Let the atomic number of the emitted particle be \(y\). Then \(116=y + 114\), so \(y=116 - 114 = 2\).

An alpha particle (\(\alpha\)) has a mass number of \(4\) and an atomic number of \(2\) (\(_{2}^{4}\alpha\)), while a beta particle (\(\beta\)) has a mass number of \(0\) and an atomic number of \(- 1\) (\(_{-1}^{0}\beta\)).

Answer:

An alpha particle