`""_(72)^(180)X to beta to gamma to ""_(Z)^(A)X'` . Z and A are

Options

A. 69, 172
B. 172, 69
C. 180, 70
D. 182, 68

Explanation

To solve the problem, we need to analyze the nuclear decay process step by step. The reaction given is:

\[ _{72}^{180}X \to \text{beta} \to \text{gamma} \to _{Z}^{A}X' \]

1. **Identify the initial values**: The element \( X \) has an atomic number \( Z = 72 \) and a mass number \( A = 180 \).

2. **Decay by alpha emission**:
- An alpha particle (\(_{2}^{4}\text{He}\)) is emitted. This reduces the atomic number by 2 and the mass number by 4.
- New atomic number: \( Z - 2 = 72 - 2 = 70 \)
- New mass number: \( A - 4 = 180 - 4 = 176 \)

So after the alpha decay, we have:
\[ _{70}^{176}X \]

3. **Decay by beta emission**:
- A beta particle (\(_{-1}^{0}\text{e}\)) is emitted. This increases the atomic number by 1 (since a neutron is converted into a proton) but does not change the mass number.
- New atomic number: \( Z + 1 = 70 + 1 = 71 \)
- Mass number remains the same: \( A = 176 \)

So after the beta decay, we have:
\[ _{71}^{176}X \]

4. **Decay by gamma emission**:
- Gamma emission does not change the atomic number or the mass number. Therefore, the values remain the same:
- Atomic number: \( Z = 71 \)
- Mass number: \( A = 176 \)

5. **Final values**: The final values for \( Z \) and \( A \) in the reaction \( _{Z}^{A}X' \) are:
\[ Z = 71, \quad A = 176 \]

Thus, the answer is:
\[ Z = 71, \quad A = 176 \]