A nucleus undergoes a series of decay according to the scheme `A overset(alpha)(rarr) B overset(beta)(rarr)C overset(alpha)(rarr)D overset(gamma)(rarr) E`
Atomic number and mass number of E are 69 and 172

Options

A. Atomic number of A is 72
B. Mass number of B is 176
C. Atomic number of D is 69
D. Atomic number of C is 69

Explanation

To solve the problem, we need to analyze the decay series step by step, keeping track of the changes in atomic number and mass number at each stage of the decay process.

### Step-by-Step Solution:

1. **Understanding the Decay Process**:
- The decay series is given as:
\( A \overset{\alpha}{\rightarrow} B \overset{\beta}{\rightarrow} C \overset{\alpha}{\rightarrow} D \overset{\gamma}{\rightarrow} E \)
- We know that:
- Alpha decay (\(\alpha\)) decreases the mass number by 4 and the atomic number by 2.
- Beta decay (\(\beta\)) does not change the mass number but decreases the atomic number by 1.
- Gamma decay (\(\gamma\)) does not change either the mass number or the atomic number.

2. **Given Information**:
- The final nucleus \( E \) has an atomic number of 69 and a mass number of 172.

3. **Finding the Properties of D**:
- Since \( E \) is reached from \( D \) through gamma decay, the properties of \( D \) are the same as those of \( E \):
- Atomic number of \( D = 69 \)
- Mass number of \( D = 172 \)

4. **Finding the Properties of C**:
- \( D \) decays to \( C \) through alpha decay:
- For alpha decay:
- Mass number decreases by 4: \( 172 + 4 = 176 \)
- Atomic number decreases by 2: \( 69 + 2 = 71 \)
- Therefore, the properties of \( C \) are:
- Atomic number of \( C = 71 \)
- Mass number of \( C = 176 \)

5. **Finding the Properties of B**:
- \( C \) decays to \( B \) through beta decay:
- For beta decay:
- Mass number remains the same: \( 176 \)
- Atomic number decreases by 1: \( 71 - 1 = 70 \)
- Therefore, the properties of \( B \) are:
- Atomic number of \( B = 70 \)
- Mass number of \( B = 176 \)

6. **Finding the Properties of A**:
- \( B \) decays to \( A \) through alpha decay:
- For alpha decay:
- Mass number decreases by 4: \( 176 + 4 = 180 \)
- Atomic number decreases by 2: \( 70 + 2 = 72 \)
- Therefore, the properties of \( A \) are:
- Atomic number of \( A = 72 \)
- Mass number of \( A = 180 \)

### Final Answer:
- The atomic number of \( E \) is 69 and the mass number is 172, which is consistent with the given data.