If `._(92)U^(238)` changes to `._(85)At^(210)` by a series of `alpha`-and `beta` -decays, the number of `alpha` and `beta`-decays undergone is .

Options

A. 7 and 5
B. 7 and 7
C. 5 and 7
D. 7 and 9

Answer

To solve the problem of how many alpha and beta decays occur when Uranium-238 (U-238) transforms into Astatine-210 (At-210), we can follow these steps:

### Step 1: Write down the decay equation
We start with the initial isotope and the final isotope:
- Initial: \( _{92}^{238}U \)
- Final: \( _{85}^{210}At \)

### Step 2: Define the decay particles
- An alpha particle can be represented as \( _{2}^{4}He \) or \( _{2}^{4}\alpha \).
- A beta particle can be represented as \( _{-1}^{0}e \) or \( _{-1}^{0}\beta \).

### Step 3: Set up the equations for mass number and atomic number
Let:
- \( n \) = number of alpha decays
- \( m \) = number of beta decays

From the decay process, we can write two equations based on the conservation of mass number and atomic number:

1. **Mass Number Equation**:
\[ 238 - 4n + 0m = 210 \]
Simplifying this gives:
\[ 238 - 210 = 4n \implies 28 = 4n \implies n = \frac{28}{4} = 7 \]

2. **Atomic Number Equation**:
\[ 92 - 2n + (-1)m = 85 \]
Simplifying this gives:
\[ 92 - 85 = 2n - m \implies 7 = 2n - m \]
Substituting \( n = 7 \) into the equation:
\[ 7 = 2(7) - m \implies 7 = 14 - m \implies m = 14 - 7 = 7 \]

### Step 4: Conclusion
From our calculations, we find:
- The number of alpha decays \( n = 7 \)
- The number of beta decays \( m = 7 \)

Thus, the final answer is:
- Number of alpha decays: 7
- Number of beta decays: 7

### Final Answer:
The number of alpha and beta decays undergone is **7 and 7**.

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