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Chapter 29: Problem 21

Identify the daughter nuclide when \(_{19}^{40} \mathrm{K}\) decays via \(\beta^{-}\) decay.

Short Answer

Expert verified
Answer: \(_{20}^{40}\mathrm{Ca}\)

Step by step solution

01

Understand β⁻ decay process

In \(\beta^{-}\) decay, a neutron in a nucleus is converted into a proton, and an electron (called a beta particle) is emitted. As a result of this decay mode, the atomic number of the nucleus increases by one, but the mass number remains the same because the number of nucleons (protons + neutrons) in the nucleus is unchanged.
02

Determine the initial atomic number and mass number

The given nuclide, \(_{19}^{40} \mathrm{K}\), has an atomic number (Z) of 19 and a mass number (A) of 40.
03

Calculate the new atomic number and mass number after β⁻ decay

In β⁻ decay, the atomic number increases by one due to the conversion of a neutron into a proton, and the mass number remains the same. Therefore, after β⁻ decay, the new atomic number will be 19 + 1 = 20, and the mass number will still be 40.
04

Identify the daughter nuclide

The resulting daughter nuclide after β⁻ decay has an atomic number of 20 and a mass number of 40, which corresponds to the element calcium (Ca). Therefore, the daughter nuclide is \(_{20}^{40}\mathrm{Ca}\).

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