Answer
Balanced nuclear equation:
$ ^{222}_{86}Rn -- \gt ^{218}_{84}Po+^4_2He$
Work Step by Step
Alpha decay:
1. Identify the reaction.
- In an $\alpha$ decay, the initial atom emits a particle with 2 neutrons and 2 protons $(_2^4He)$.
2. Determine the atomic number and symbol for : $Radon$, and write its complete symbol.
- 86, Rn. Therefore, the symbol for Radon-222 is: $^{222}_{86}Rn$.
3. Now, subtract 4 from the mass number, and 2 from the atomic number:
$^{222}_{86}Rn --\gt ^{218}_{84}?$
4. Identify the symbol for the element with atomic number equal to 84.
- It is "Po". So, the main product of that reaction will be: $^{218}_{84}Po$
5. Write the complete equation:
$ ^{222}_{86}Rn -- \gt ^{218}_{84}Po+^4_2He$
