Question

complete and balance each of the following nuclear equations.
part 1 of 2
$square \
ightarrow \ce{_{-1}^0 e} + \ce{_{83}^{210} bi}$
part 2 of 2
$ce{_{11}^{22} na} \
ightarrow \square + \ce{_{10}^{22} ne}$

Explanation:

Part 1:

Step1: Balance atomic number

The atomic number on the right - hand side: \(- 1+83 = 82\). So the atomic number of the missing element is \(82\).

Step2: Balance mass number

The mass number on the right - hand side: \(0 + 210=210\). So the mass number of the missing element is \(210\).
The element with atomic number \(82\) is \(Pb\) (lead). So the missing nuclide is \(_{82}^{210}Pb\).

Part 2:

Step1: Balance atomic number

The atomic number on the left - hand side is \(11\), and on the right - hand side is \(10\). So the atomic number of the missing particle is \(11-10 = 1\).

Step2: Balance mass number

The mass number on the left - hand side is \(22\), and on the right - hand side is \(22\). So the mass number of the missing particle is \(0\).
The particle with atomic number \(1\) and mass number \(0\) is \(_{1}^{0}e\) (positron). So the missing nuclide is \(_{1}^{0}e\).

Answer:

Part 1: \(_{82}^{210}Pb\)
Part 2: \(_{1}^{0}e\)