What is the daughter nucleus (nuclide) produced when 6ªCu undergoes beta decay by emitting an electron? Replace each question mark with the appropriate integer or symbol.
What is the daughter nucleus (nuclide) produced when 6ªCu undergoes beta decay by emitting an electron? Replace each question mark with the appropriate integer or symbol.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Beta Decay of Copper-64 (Cu-64)
**Question:**
What is the daughter nucleus (nuclide) produced when \(^64\text{Cu}\) undergoes beta decay by emitting an electron? Replace each question mark with the appropriate integer or symbol.
**Answer:**
**Daughter nucleus (nuclide):**
\[
^{?}_{?}\textbf{?}
\]
### Explanation:
When \(^64\text{Cu}\) undergoes beta decay, an electron (\(e^-\)) is emitted, and a neutron in the nucleus is transformed into a proton. This process increases the atomic number by one while keeping the mass number the same.
In general, the reaction can be represented as:
\[
^{A}_{Z}\text{X} \rightarrow ^{A}_{Z+1}\text{Y} + e^-
\]
where \(A\) is the mass number (remains unchanged), and \(Z\) is the atomic number (increases by 1).
For \(^64\text{Cu}\):
- Mass number \(A = 64\)
- Atomic number of Copper \(Z = 29\)
After beta decay:
- New atomic number \(Z' = 29 + 1 = 30\)
- The element with atomic number 30 is Zinc (Zn)
Thus, the daughter nucleus produced is \(^64\text{Zn}\).
**Final Answer:**
\[
^{64}_{30}\textbf{Zn}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7d591113-c73d-4bf8-a1b7-9966513ce440%2Fdc8a4764-7ed4-4a34-9614-9356b1b8f8a8%2F11rkox_processed.png&w=3840&q=75)
Transcribed Image Text:### Beta Decay of Copper-64 (Cu-64)
**Question:**
What is the daughter nucleus (nuclide) produced when \(^64\text{Cu}\) undergoes beta decay by emitting an electron? Replace each question mark with the appropriate integer or symbol.
**Answer:**
**Daughter nucleus (nuclide):**
\[
^{?}_{?}\textbf{?}
\]
### Explanation:
When \(^64\text{Cu}\) undergoes beta decay, an electron (\(e^-\)) is emitted, and a neutron in the nucleus is transformed into a proton. This process increases the atomic number by one while keeping the mass number the same.
In general, the reaction can be represented as:
\[
^{A}_{Z}\text{X} \rightarrow ^{A}_{Z+1}\text{Y} + e^-
\]
where \(A\) is the mass number (remains unchanged), and \(Z\) is the atomic number (increases by 1).
For \(^64\text{Cu}\):
- Mass number \(A = 64\)
- Atomic number of Copper \(Z = 29\)
After beta decay:
- New atomic number \(Z' = 29 + 1 = 30\)
- The element with atomic number 30 is Zinc (Zn)
Thus, the daughter nucleus produced is \(^64\text{Zn}\).
**Final Answer:**
\[
^{64}_{30}\textbf{Zn}
\]
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