Answer
Balanced nuclear reaction:
$^{80}_{38}Sr -- \gt ^{80}_{37}Rb + {_+}_{1}^0e$
Work Step by Step
1. Identify the reaction.
- In a positron emission, one proton from the initial atom breaks into a neutron and a positron. The neutron remains in the nucleus, and the positron is released. We represent that positron by: $({_+}_{1}^0e)$.
2. Determine the atomic number and symbol for : $strontium$, and write its complete symbol.
- 38 and "Sr". Therefore, the symbol for strontium-80 is: $^{80}_{38}Sr$.
3. Now, subtract 1 from the atomic number of that strontium:
$^{80}_{38}Sr --\gt ^{80}_{37}?$
4. Identify the symbol for the element with atomic number equal to 37.
- It is "Rb". So, the main product of that reaction will be: $^{80}_{37}Rb$
5. Write the complete equation:
$^{80}_{38}Sr -- \gt ^{80}_{37}Rb + {_+}_{1}^0e$
