Nuclear equations show the process that happens when a nucleus decays. Balancing a nuclear equation is very similar to balancing a chemical equation except, that as change occur in the nucleus, the atomic symbol may change. When balancing a nuclear equation:
- make sure that the mass number is the same before the decay and after the decay: the mass number of the nucleus before decay equals the sum of the mass number of the nucleus after decay plus the mass number of the emitted particle,
- make sure that the charge is the same before and after the decay: the atomic number of the nucleus before decay equals the sum of the atomic number of the nucleus after decay plus the charge of the emitted particle and
- for α, β+ and β- decay, the atomic number changes in step (2) so the atomic symbol will change too: the element changes!
An α particle has a mass of 4 atomic units and a charge of +2 atomic units. When an α particle is emitted:
- the mass of the nucleus decreases by 4 units,
- the atomic number of the nucleus decreases by 2 units and so
- the atomic symbol of the element changes: the atomic symbol of the new element is that of the element 2 spaces to the left of the original element on the Periodic Table.
For example:
| 212 | Bi | → | 208 | Tl | + | 4 | α |
| Mass before decay = 212. Mass after decay = 208 + 4 = 212. |
| 83 | 81 | 2 |
| Charge before decay = 83. Charge after decay = 81 + 2 = 83. The element with atomic number = 81 is Tl. |
| 238 | U | → | 234 | Th | + | 4 | α |
| Mass before decay = 238. Mass after decay = 234 + 4 = 238. |
| 92 | 90 | 2 |
| Charge before decay = 92. Charge after decay = 90 + 2 = 92. The element with atomic number = 90 is Th. |
| 234 | U | → | 230 | Th | + | 4 | α |
| Mass before decay = 234. Mass after decay = 230 + 4 = 234. |
| 92 | 90 | 2 |
| Charge before decay = 92. Charge after decay = 90 + 2 = 92. The element with atomic number = 90 is Th. |
| 230 | Th | → | 226 | Ra | + | 4 | α |
| Mass before decay = 230. Mass after decay = 226 + 4 = 230. |
| 90 | 88 | 2 |
| Charge before decay = 90. Charge after decay = 88 + 2 = 90. The element with atomic number = 88 is Ra. |
A β
+ particle has effectively zero mass and a charge of +1 atomic units. When a β
+ particle is emitted:
- the mass of the nucleus does not change,
- the atomic number of the nucleus decreases by 1 unit and so
- the atomic symbol of the element changes: the atomic symbol of the new element is that of the element 1 space to the left of the original element on the Periodic Table.
For example:
| 12 | N | → | 12 | C | + | 0 | β |
| Mass before decay = 12. Mass after decay = 12 + 0 = 12. |
| 7 | 6 | +1 |
| Charge before decay = 7. Charge after decay = 6 + 1 = 7. The element with atomic number = 6 is C. |
| 11 | C | → | 11 | B | + | 0 | β |
| Mass before decay = 11. Mass after decay = 11 + 0 = 11. |
| 6 | 5 | +1 |
| Charge before decay = 6. Charge after decay = 5 + 1 = 6. The element with atomic number = 5 is B. |
| 23 | Mg | → | 23 | Na | + | 0 | β |
| Mass before decay = 23. Mass after decay = 23 + 0 = 23. |
| 12 | 11 | +1 |
| Charge before decay = 12. Charge after decay = 11 + 1 = 12. The element with atomic number = 11 is Na. |
| 15 | O | → | 15 | N | + | 0 | β |
| Mass before decay = 15. Mass after decay = 15 + 0 = 15. |
| 8 | 7 | +1 |
| Charge before decay = 8. Charge after decay = 7 + 1 = 8. The element with atomic number = 7 is N. |
A β
+ particle has effectively zero mass and a charge of -1 atomic units. When a β
- particle is emitted:
- the mass of the nucleus does not change,
- the atomic number of the nucleus increases by 1 unit and so
- the atomic symbol of the element changes: the atomic symbol of the new element is that of the element 1 space to the right of the original element on the Periodic Table.
For example:
| 12 | B | → | 12 | C | + | 0 | β |
| Mass before decay = 12. Mass after decay = 12 + 0 = 12. |
| 5 | 6 | -1 |
| Charge before decay = 5. Charge after decay = 6 + (-1) = 5. The element with atomic number = 6 is C. |
| 14 | C | → | 14 | N | + | 0 | β |
| Mass before decay = 14. Mass after decay = 14 + 0 = 14. |
| 6 | 7 | -1 |
| Charge before decay = 6. Charge after decay = 7 + (-1) = 6. The element with atomic number = 7 is N. |
| 15 | C | → | 15 | N | + | 0 | β |
| Mass before decay = 15. Mass after decay = 15 + 0 = 15. |
| 6 | 7 | -1 |
| Charge before decay = 6. Charge after decay = 7 + (-1) = 6. The element with atomic number = 7 is N. |
| 137 | Cs | → | 137 | Ba | + | 0 | β |
| Mass before decay = 137. Mass after decay = 137 + 0 = 137. |
| 11 | 12 | -1 |
| Charge before decay = 11. Charge after decay = 12 + (-1) = 11. The element with atomic number = 12 is Ba. |
Each quiz is made up of 10 questions randomly drawn from a large set so you can repeat the quiz many times over. The results are not stored!
Multiple choice practice questions (not an assessment)
*If you are unsure how to extract the number of protons and neutraons from the atomic symbol, review the '
atomic symbols tutorial.
Want to be able to predict which nuclei are radioactive and whether they will decay by α, β
+ or β
-? Check the '
predicting nuclear decay' iChem resource.
