Question

what is the missing particle?

$\

$$\begin{array}{r} 99 \\\\ 43 \\end{array}$$

\text{tc} \
ightarrow \square + \

$$\begin{array}{r} 99 \\\\ 44 \\end{array}$$

\text{ru}$

Explanation:

Step1: Balance mass number

Mass number of \(^{99}_{43}Tc\) is \(99\), mass number of \(^{99}_{44}Ru\) is \(99\). Let the mass number of the missing particle be \(A\). Then \(99=A + 99\), so \(A = 0\).

Step2: Balance atomic number

Atomic number of \(^{99}_{43}Tc\) is \(43\), atomic number of \(^{99}_{44}Ru\) is \(44\). Let the atomic number of the missing particle be \(Z\). Then \(43=Z + 44\), so \(Z=-1\).

Step3: Identify the particle

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

Answer:

\(^{\ 0}_{-1}e\) (the third option)