Answer
Balanced nuclear equation:
$^{210}_{88}Ra -- \gt ^{206}_{86}Rn + ^4_2He$
Work Step by Step
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 for : $Ra$, and write its complete symbol.
- It is 88. Therefore, the symbol for Ra-210 is: $^{210}_{88}Ra$.
3. Now, subtract 4 from the mass number, and 2 from the atomic number:
$^{210}_{88}Ra --\gt ^{206}_{86}?$
4. Identify the symbol for the element with atomic number equal to 86.
- It is "Rn". So, the main product of that reaction will be: $^{206}_{86}Rn$
5. Write the complete equation:
$^{210}_{88}Ra -- \gt ^{206}_{86}Rn + ^4_2He$
