Answer
Balanced nuclear equation:
$^{137}_{55}Cs -- \gt ^{137}_{56}Ba + ^{0}_{-1}e$
Work Step by Step
1. Identify the reaction.
- In a $\beta$ decay, one neutron from the initial atom breaks into one proton and one electron. The proton remains in the nucleus, and the electron is released. We represent that electron by: $(_{-1}^0e)$.
2. Determine the atomic number for : $cesium$, and write its complete symbol.
- It is 55. Therefore, the symbol for cesium-137 is: $^{137}_{55}Cs$.
3. Now, add 1 to the atomic number of the cesium:
$^{137}_{55}Cs --\gt ^{137}_{56}?$
4. Identify the symbol for the element with atomic number equal to 56.
- It is "Ba". So, the main product of that reaction will be: $^{137}_{56}Ba$
5. Write the complete equation:
$^{137}_{55}Cs -- \gt ^{137}_{56}Ba + ^{0}_{-1}e$
