`._(92)^(235)U` decays with emission of `alpha` and `beta`- particles to form ultimately `._(82)^(207)Pb`. How many `alpha` and `beta`-particles are emitted per atom of Pb produced?

Answer

To solve the problem of how many alpha and beta particles are emitted when Uranium-235 decays to form Lead-207, we can follow these steps:

### Step 1: Understand the decay process
- Uranium-235 (U-235) has an atomic number (Z) of 92 and a mass number (A) of 235.
- Lead-207 (Pb-207) has an atomic number (Z) of 82 and a mass number (A) of 207.
- During the decay, U-235 emits alpha (α) and beta (β) particles.

### Step 2: Set up the equations
- Let \( n \) be the number of alpha particles emitted.
- Let \( x \) be the number of beta particles emitted.

The decay can be represented as:
\[ _{92}^{235}U \rightarrow \, _{82}^{207}Pb + n \, _{2}^{4}\alpha + x \, _{0}^{-1}\beta \]

### Step 3: Balance the mass numbers
The total mass number before and after the decay must be equal:
\[ 235 = 207 + 4n + 0x \]
This simplifies to:
\[ 235 = 207 + 4n \]
Rearranging gives:
\[ 4n = 235 - 207 \]
\[ 4n = 28 \]
\[ n = \frac{28}{4} = 7 \]

### Step 4: Balance the atomic numbers
The total atomic number before and after the decay must also be equal:
\[ 92 = 82 + 2n - x \]
Substituting \( n = 7 \):
\[ 92 = 82 + 2(7) - x \]
This simplifies to:
\[ 92 = 82 + 14 - x \]
Rearranging gives:
\[ 92 = 96 - x \]
\[ x = 96 - 92 \]
\[ x = 4 \]

### Step 5: Conclusion
- The number of alpha particles emitted is \( n = 7 \).
- The number of beta particles emitted is \( x = 4 \).

Thus, for each atom of Lead-207 produced from the decay of Uranium-235, **7 alpha particles and 4 beta particles are emitted**.

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