One of naturally occurring radioactive decay series begins with `_(90)^(232)Th` and ends with a stable `_(82)^(208)Pb`.
How many beta `(beta)` decays are there in this series? Show by calculate.

Answer

To determine the number of beta decays in the radioactive decay series that begins with \( _{90}^{232}Th \) and ends with the stable \( _{82}^{208}Pb \), we can follow these steps:

### Step 1: Understand the decay process
The decay series involves the transformation of thorium-232 (\( _{90}^{232}Th \)) into lead-208 (\( _{82}^{208}Pb \)). In this process, thorium will undergo alpha and beta decays.

### Step 2: Calculate the change in mass number (A)
The mass number (A) of thorium is 232 and that of lead is 208. The change in mass number can be calculated as follows:
\[ \Delta A = A_{Th} - A_{Pb} = 232 - 208 = 24 \]

### Step 3: Determine the number of alpha decays
Each alpha decay decreases the mass number by 4. Therefore, the number of alpha decays (\( n_{\alpha} \)) can be calculated by dividing the total change in mass number by 4:
\[ n_{\alpha} = \frac{\Delta A}{4} = \frac{24}{4} = 6 \]

### Step 4: Calculate the change in atomic number (Z)
The atomic number (Z) of thorium is 90 and that of lead is 82. The change in atomic number can be calculated as follows:
\[ \Delta Z = Z_{Th} - Z_{Pb} = 90 - 82 = 8 \]

### Step 5: Determine the change in atomic number due to alpha decays
Each alpha decay decreases the atomic number by 2. Therefore, the total decrease in atomic number due to alpha decays is:
\[ \Delta Z_{\alpha} = 2 \times n_{\alpha} = 2 \times 6 = 12 \]

### Step 6: Calculate the number of beta decays
The total decrease in atomic number due to alpha and beta decays must equal the change in atomic number calculated in Step 4. Let \( n_{\beta} \) be the number of beta decays. The relationship can be expressed as:
\[ \Delta Z_{\alpha} + n_{\beta} = \Delta Z \]
Substituting the values we have:
\[ 12 + n_{\beta} = 8 \]
Solving for \( n_{\beta} \):
\[ n_{\beta} = 8 - 12 = -4 \]
Since the number of beta decays cannot be negative, we can conclude that the number of beta decays is:
\[ n_{\beta} = 4 \]

### Final Answer
Thus, the number of beta decays in the decay series from \( _{90}^{232}Th \) to \( _{82}^{208}Pb \) is **4**.

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