Question

What is the daughter nucleus (nuclide) produced when 64cu undergoes beta decay by emitting an electron? replace each question mark with the appropriate integer or symbol?

Answer 1

The daughter nucleus produced when 64Cu undergoes beta decay is 64Zn (Zinc-64), resulting from the conversion of a neutron into a proton within the nucleus.

Step-by-step explanation:

When 64Cu (Copper-64) undergoes beta decay by emitting an electron, the daughter nucleus, or nuclide, produced is 64Zn (Zinc-64). This is because in beta decay, a neutron in the nucleus is converted into a proton while releasing an electron (beta particle) and an antineutrino. The atomic number increases by one, but the mass number remains unchanged. Therefore, the beta decay equation for Cu-64 is:

64Cu→ 64Zn + β˃ + √e

Where β˃ represents the emitted beta particle (electron), and √e is the antineutrino. This transformation results in the new nuclide, Zinc-64, which has one more proton than Copper-64.

Answer 2

Answer is: daughter nucleus is zinc.
Nuclear reaction 3: ⁶⁴Cu → ⁶⁴Zn + e⁻ + electron antineutrino.
Beta decay is radioactive decay in which a beta ray and a neutrino are emitted from an atomic nucleus.
There are two types of beta decay: beta minus and beta plus.
In beta minus decay, neutron is converted to a proton and an electron and an electron antineutrino and atomic number Z is increased by one.