The energy E of a particle varies with time t according to the equation `E=E_0sin(alphat).e^((-alphat)/(betax))`, where x is displacement from mean position `E_0` is energy at infinite position and `alpha` and `beta` are constants .
Dimensions of `beta` are

Options

A. `[M^(-1) L^0 T^0]`
B. `[M^0 L^(-1) T^0]`
C. `[M^0 L^0 T^(-1)]`
D. `[M^0 L^0 T^0]`

Answer

To find the dimensions of the constant \(\beta\) in the given equation for energy \(E\), we start with the equation:

\[ E = E_0 \sin(\alpha t) \cdot e^{-\frac{\alpha t}{\beta x}} \]

### Step 1: Understand the components of the equation

The equation consists of three parts: \(E_0\), \(\sin(\alpha t)\), and \(e^{-\frac{\alpha t}{\beta x}}\).

1. **Energy \(E\)** has dimensions of energy, which is \([E] = [M L^2 T^{-2}]\).
2. **\(E_0\)** is also energy, so it has the same dimensions as \(E\).
3. **\(\sin(\alpha t)\)** is dimensionless, meaning \(\alpha t\) must also be dimensionless.
4. **The term \(e^{-\frac{\alpha t}{\beta x}}\)** must also be dimensionless.

### Step 2: Analyze the term \(\frac{\alpha t}{\beta x}\)

Since both \(\alpha t\) and \(\beta x\) must be dimensionless, we can equate their dimensions:

\[ [\alpha t] = [\beta x] \]

### Step 3: Find the dimensions of \(\alpha t\)

The term \(\alpha t\) involves \(\alpha\) and time \(t\). Since \(t\) has dimensions of time \([T]\), we can express the dimensions of \(\alpha\):

\[ [\alpha t] = [\alpha][T] \]

For \(\alpha t\) to be dimensionless, we have:

\[ [\alpha][T] = [M^0 L^0 T^0] \Rightarrow [\alpha] = [T^{-1}] \]

### Step 4: Find the dimensions of \(\beta x\)

Next, we analyze \(\beta x\). The displacement \(x\) has dimensions of length \([L]\), so we can express the dimensions of \(\beta\):

\[ [\beta x] = [\beta][L] \]

For \(\beta x\) to be dimensionless, we have:

\[ [\beta][L] = [M^0 L^0 T^0] \Rightarrow [\beta] = [L^{-1}] \]

### Step 5: Final dimensions of \(\beta\)

Since we found that \(\beta\) has dimensions of \([L^{-1}]\), we can express it in terms of mass and time as follows:

\[ [\beta] = [M^0 L^{-1} T^0] \]

Thus, the final dimensions of \(\beta\) are:

\[ \text{Dimensions of } \beta = [M^0 L^{-1} T^0] \]

### Conclusion

The dimensions of \(\beta\) are \(L^{-1}\), which corresponds to option B.

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