suppose a beryllium - 10 nuclide decays into a boron - 10 nuclide by emitting an electron. complete the…
suppose a beryllium - 10 nuclide decays into a boron - 10 nuclide by emitting an electron. complete the nuclear chemical equation below so that it describes this nuclear reaction.\n$_{4}^{10}be\rightarrow\\square$
Answer
Explanation:
Step1: Recall beta - decay rules
In beta - minus decay, a neutron in the nucleus is converted into a proton, an electron, and an antineutrino. The mass number ($A$) remains the same and the atomic number ($Z$) increases by 1.
Step2: Identify initial values
For beryllium - 10 ($^{10}_{4}Be$), the mass number $A = 10$ and the atomic number $Z = 4$.
Step3: Determine products
Since the mass number remains 10 and the atomic number increases by 1 (from 4 to 5), the resulting nucleus is boron - 10 ($^{10}{5}B$). Also, an electron (${ - 1}^{0}e$) and an antineutrino ($\overline{\nu}e$) are emitted. The nuclear equation is $^{10}{4}Be\rightarrow^{10}_{5}B + _{ - 1}^{0}e+\overline{\nu}_e$.
Answer:
$^{10}_{5}B + _{ - 1}^{0}e+\overline{\nu}_e$