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Chapter 29: Problem 23

Write out the reaction and identify the daughter nuclide when \(_{11}^{22}\) Na decays by electron capture.

Short Answer

Expert verified
Answer: When _{11}^{22}Na undergoes electron capture decay, the daughter nuclide produced is _{10}^{22}Ne. The reaction for this decay process is: _{11}^{22}Na + e^{-} → _{10}^{22}Ne + ν_e.

Step by step solution

01

Write the electron capture decay reaction

In an electron capture decay, the nucleus captures an electron which combines with a proton to form a neutron. The general form of this decay reaction for an element X can be written as: \[_{Z}^{A}\textrm{X} + e^{-} \rightarrow _{Z - 1}^{A}\textrm{Y} + \nu_e\] Here \(_{Z}^{A}\textrm{X}\) represents the parent nucleus, \(e^{-}\) represents the electron being captured, \(_{Z - 1}^{A}\textrm{Y}\) is the daughter nucleus produced, and \(\nu_e\) is an emitted neutrino.
02

Apply the reaction to the given isotope: \(_{11}^{22}\)Na

We will apply the electron capture decay to the given isotope. The initial nucleus is \(_{11}^{22}\textrm{Na}\), so the atomic number Z is 11, and the mass number A is 22: \[_{11}^{22}\textrm{Na} + e^{-} \rightarrow _{11 - 1}^{22}\textrm{Y} + \nu_e\] \[_{11}^{22}\textrm{Na} + e^{-} \rightarrow _{10}^{22}\textrm{Y} + \nu_e\]
03

Identify the daughter nuclide

Now we can identify which element has an atomic number (Z) of 10 to find out the daughter nuclide. By referring to the periodic table, the element with Z=10 is neon (Ne). Thus, the daughter nuclide is \(_{10}^{22}\)Ne: \[_{11}^{22}\textrm{Na} + e^{-} \rightarrow _{10}^{22}\textrm{Ne} + \nu_e\] So, when \(_{11}^{22}\)Na decays by electron capture, the daughter nuclide formed is \(_{10}^{22}\)Ne.

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