Chapter 30, Problem 9Q

To determine

Resulting nuclide for decay of the isotope ${}_{29}{}^{64}Cu$ by γ, β^-, β⁺ emission

Answer to Problem 9Q

Solution:

The resulting nuclide for γ decay is the same isotope of copper

The resulting nuclide for β − decay is an isotope of zinc

The resulting nuclide for β + decay is an isotope of nickel

Explanation of Solution

When adecay of the isotope ${}_{29}{}^{64}Cu$ takes place by γ, β^-, β⁺ emission, the resulting nuclide will be,

The resulting nuclide for γ decay is the same isotope. Here resulting nuclide will be same but it will be in lower energy state.

${}_{29}{}^{64}Cu\to {}_{29}{}^{64}Cu+\gamma$

The resulting nuclide for β − decay is an isotope of zinc,

${}_{29}{}^{64}Cu\to {}_{30}{}^{64}Zn+e^{-}+v$

The resulting nuclide for β + decay is an isotope of nickel, 64

${}_{29}{}^{64}Cu\to {}_{28}{}^{64}Ni+e^{+}+v$