`._(83)^(214)Bi` decays to A by `alpha` -emission . A then decays to B by beta emission , which further decays to C by another beta emission . Element C decays to D by still another beta emission , and D deacays by `alpha`-emission to form a stable isotope E. What is element E?

Options

A. `_(81)^(207)T1`
B. `_(80)^(206)Hg`
C. `_(79)^(206)Au`
D. `_(82)^(206)Pb`

Answer

To solve the problem step by step, we will track the decay of each element until we reach the stable isotope E.

### Step 1: Identify the decay of Bismuth (Bi)
- The initial element is Bismuth, represented as \( _{83}^{214}Bi \).
- Bismuth decays to element A by alpha emission.
- In alpha emission, the mass number decreases by 4 and the atomic number decreases by 2.

**Calculation:**
- Mass number of A = \( 214 - 4 = 210 \)
- Atomic number of A = \( 83 - 2 = 81 \)

Thus, element A is \( _{81}^{210}Tl \) (Thallium).

### Step 2: Decay of A to B
- Element A decays to element B by beta emission.
- In beta emission, the mass number remains unchanged, and the atomic number increases by 1.

**Calculation:**
- Mass number of B = 210 (remains the same)
- Atomic number of B = \( 81 + 1 = 82 \)

Thus, element B is \( _{82}^{210}Pb \) (Lead).

### Step 3: Decay of B to C
- Element B decays to element C by another beta emission.
- Again, the mass number remains unchanged, and the atomic number increases by 1.

**Calculation:**
- Mass number of C = 210 (remains the same)
- Atomic number of C = \( 82 + 1 = 83 \)

Thus, element C is \( _{83}^{210}Bi \) (Bismuth).

### Step 4: Decay of C to D
- Element C decays to element D by another beta emission.
- The mass number remains unchanged, and the atomic number increases by 1.

**Calculation:**
- Mass number of D = 210 (remains the same)
- Atomic number of D = \( 83 + 1 = 84 \)

Thus, element D is \( _{84}^{210}Po \) (Polonium).

### Step 5: Decay of D to E
- Element D decays to element E by alpha emission.
- In alpha emission, the mass number decreases by 4 and the atomic number decreases by 2.

**Calculation:**
- Mass number of E = \( 210 - 4 = 206 \)
- Atomic number of E = \( 84 - 2 = 82 \)

Thus, element E is \( _{82}^{206}Pb \) (Lead).

### Final Answer
Element E is \( _{82}^{206}Pb \) (Lead).

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