The time dependence of a physical quantity P is given by `P=P_(0) exp (-alpha t^(2))`, where `alpha` is a constant and t is time. The constant `alpha`

Options

A. [T]
B. `[T^2]`
C. `[T^(-1)]`
D. `[T^(-2)]`

Answer

To find the dimensions of the constant \( \alpha \) in the equation \( P = P_0 e^{-\alpha t^2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Equation**:
The given equation is \( P = P_0 e^{-\alpha t^2} \). Here, \( P \) is a physical quantity that depends on time \( t \), and \( P_0 \) is a constant. The term \( e^{-\alpha t^2} \) must be dimensionless since the exponent of an exponential function must not have any dimensions.

2. **Identify the Dimensions**:
Since \( e^{-\alpha t^2} \) is dimensionless, we can write:
\[ -\alpha t^2 \text{ must be dimensionless.} \]
This implies that the dimensions of \( \alpha t^2 \) must cancel out to give a dimensionless quantity.

3. **Dimensions of Time**:
The dimension of time \( t \) is given as:
\[ [t] = T. \]
Therefore, the dimensions of \( t^2 \) are:
\[ [t^2] = T^2. \]

4. **Setting Up the Equation**:
Since \( \alpha t^2 \) is dimensionless, we can express this as:
\[ [\alpha] \cdot [t^2] = 1. \]
Substituting the dimensions of \( t^2 \):
\[ [\alpha] \cdot T^2 = 1. \]

5. **Solving for Dimensions of \( \alpha \)**:
Rearranging the equation gives:
\[ [\alpha] = \frac{1}{[t^2]} = T^{-2}. \]

6. **Final Dimensions**:
Since \( \alpha \) does not depend on length \( L \) or mass \( M \), we can express the dimensions of \( \alpha \) as:
\[ [\alpha] = L^0 M^0 T^{-2} = T^{-2}. \]

### Conclusion:
The dimensional formula for \( \alpha \) is \( L^0 M^0 T^{-2} \) or simply \( T^{-2} \).