`._72Re^162 to._(z_1)X^(A_1)+` alpha particle
`._74W^(188)to._(Z_2)Y^(A_2)`+Beta particle
value of `(A_2-A_1)/(Z_2-Z_1)` is :

Answer

To solve the problem, we need to analyze the two nuclear reactions given and extract the values of \(A_1\), \(A_2\), \(Z_1\), and \(Z_2\).

### Step 1: Analyze the first reaction
The first reaction is:
\[ ^{162}_{72}Re \rightarrow ^{A_1}_{Z_1}X + \text{alpha particle} \]

An alpha particle has a mass number of 4 and an atomic number of 2. Therefore, we can write:
\[ ^{162}_{72}Re \rightarrow ^{A_1}_{Z_1}X + ^{4}_{2}He \]

### Step 2: Calculate \(Z_1\) and \(A_1\)
Using the conservation of atomic number and mass number:
1. **For mass number (A)**:
\[ 162 = A_1 + 4 \implies A_1 = 162 - 4 = 158 \]

2. **For atomic number (Z)**:
\[ 72 = Z_1 + 2 \implies Z_1 = 72 - 2 = 70 \]

So, we have:
- \(A_1 = 158\)
- \(Z_1 = 70\)

### Step 3: Analyze the second reaction
The second reaction is:
\[ ^{188}_{74}W \rightarrow ^{A_2}_{Z_2}Y + \text{beta particle} \]

A beta particle (electron) has a mass number of 0 and an atomic number of -1. Therefore, we can write:
\[ ^{188}_{74}W \rightarrow ^{A_2}_{Z_2}Y + ^{0}_{-1}e \]

### Step 4: Calculate \(Z_2\) and \(A_2\)
Using the conservation of atomic number and mass number:
1. **For mass number (A)**:
\[ 188 = A_2 + 0 \implies A_2 = 188 \]

2. **For atomic number (Z)**:
\[ 74 = Z_2 - 1 \implies Z_2 = 74 + 1 = 75 \]

So, we have:
- \(A_2 = 188\)
- \(Z_2 = 75\)

### Step 5: Calculate the ratio \((A_2 - A_1) / (Z_2 - Z_1)\)
Now we can substitute the values into the ratio:
\[ \frac{A_2 - A_1}{Z_2 - Z_1} = \frac{188 - 158}{75 - 70} \]

Calculating the numerator and denominator:
- Numerator: \(188 - 158 = 30\)
- Denominator: \(75 - 70 = 5\)

Thus, the ratio is:
\[ \frac{30}{5} = 6 \]

### Final Answer
The value of \(\frac{A_2 - A_1}{Z_2 - Z_1}\) is \(6\).

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