A radioactive element `ThA( ._(84)Po^(216))` can undergo `alpha` and `beta` are type of disintegrations with half-lives, `T_1` and `T_2` respectively. Then the half-life of ThA is

Options

A. `T_(1)+T_(2)`
B. `T_(1)T_(2)`
C. `T_(1)-T_(2)`
D. `(T_(1)T_(2))/(T_(1)+T_(2))`

Explanation

To find the half-life of the radioactive element ThA (Polonium-216) that can undergo both alpha and beta disintegrations, we can use the relationship between the decay constants and half-lives of the processes involved.

### Step-by-Step Solution:

1. **Understand the Decay Processes**:
- The element can undergo two types of decay: alpha decay and beta decay.
- Let the half-life for alpha decay be \( T_1 \) and for beta decay be \( T_2 \).

2. **Define Decay Constants**:
- The decay constant \( \lambda_1 \) for alpha decay is related to its half-life \( T_1 \) by the formula:
\[ \lambda_1 = \frac{0.693}{T_1} \]
- Similarly, the decay constant \( \lambda_2 \) for beta decay is:
\[ \lambda_2 = \frac{0.693}{T_2} \]

3. **Total Decay Constant**:
- The total decay constant \( \lambda \) for the element is the sum of the individual decay constants:
\[ \lambda = \lambda_1 + \lambda_2 \]
- Substituting the expressions for \( \lambda_1 \) and \( \lambda_2 \):
\[ \lambda = \frac{0.693}{T_1} + \frac{0.693}{T_2} \]

4. **Express Total Half-Life**:
- The total half-life \( T \) is related to the total decay constant by:
\[ \lambda = \frac{0.693}{T} \]
- Setting the two expressions for \( \lambda \) equal gives:
\[ \frac{0.693}{T} = \frac{0.693}{T_1} + \frac{0.693}{T_2} \]

5. **Simplify the Equation**:
- Cancel \( 0.693 \) from both sides:
\[ \frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2} \]

6. **Find Total Half-Life**:
- Rearranging the equation gives:
\[ \frac{1}{T} = \frac{T_1 + T_2}{T_1 T_2} \]
- Taking the reciprocal of both sides results in:
\[ T = \frac{T_1 T_2}{T_1 + T_2} \]

### Final Answer:
The half-life of the radioactive element ThA (Polonium-216) is:
\[ T = \frac{T_1 T_2}{T_1 + T_2} \]