Complete each of the following nuclear equations and describe the type of radiation:
c. ⁶⁶₂₉Cu → ⁶⁶₃₀Zn + ?

Guida verificata passo dopo passo

Step 1: Analyze the given nuclear equation. The reactant is ⁶⁶₂₉Cu (Copper-66), and one of the products is ⁶⁶₃₀Zn (Zinc-66). Notice that the mass number (66) remains the same, but the atomic number increases by 1 (from 29 to 30). This suggests a beta decay process.

Step 2: Recall that in beta decay, a neutron in the nucleus is converted into a proton, and a beta particle (an electron, e⁻) is emitted. This increases the atomic number by 1 while keeping the mass number constant.

Step 3: Write the missing particle in the equation. The emitted beta particle is represented as

β

^- or

e

^-, with a mass number of 0 and an atomic number of -1.

Step 4: Complete the nuclear equation:

^{66}_{29}Cu → ^{66}_{30}Zn + β

^-. This shows that Copper-66 undergoes beta decay to form Zinc-66 and a beta particle.

Step 5: Describe the type of radiation. Beta radiation involves the emission of high-energy electrons (beta particles) from the nucleus during the decay process.

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