To what stable isotope would `""_(103)^(257)Lr` decay?

Options

A. `""_(82)^(208)Pb`
B. `""_(83)^(209)Bi`
C. `""_(82)^(206)Pb`
D. `""_(82)^(207)Pb`

Explanation

To determine the stable isotope to which \( _{103}^{257}Lr \) (Lawrencium) decays, we can follow these steps:

### Step 1: Identify the Element
The notation \( _{103}^{257}Lr \) indicates that Lawrencium has an atomic number of 103 and an atomic mass of 257.

### Step 2: Determine the Decay Series
Lawrencium (Lr) belongs to the actinide series and is known to undergo alpha decay. Since it is part of the 4n + 1 series, we can use this information to predict its decay product.

### Step 3: Calculate the Decay Product
In alpha decay, an alpha particle (which consists of 2 protons and 2 neutrons) is emitted. This means that the atomic number decreases by 2 and the atomic mass decreases by 4.

- **New Atomic Number**:
\[ 103 - 2 = 101 \]

- **New Atomic Mass**:
\[ 257 - 4 = 253 \]

Thus, the decay product is \( _{101}^{253}Md \) (Mendelevium).

### Step 4: Determine Stability
Mendelevium is not stable and will continue to decay. However, we need to find the end product of this decay chain. Mendelevium also belongs to the actinide series and will eventually decay to a stable isotope.

### Step 5: Identify the Stable Isotope
The stable isotope that Mendelevium decays to is Bismuth (Bi). The atomic number of Bismuth is 83, and its most stable isotope has an atomic mass of 209.

### Conclusion
Thus, the stable isotope to which \( _{103}^{257}Lr \) decays is \( _{83}^{209}Bi \) (Bismuth).

### Final Answer
The stable isotope to which \( _{103}^{257}Lr \) decays is \( _{83}^{209}Bi \).

---