Pressure depends on distance as, $P = (\alpha/\beta)\exp(-\alpha z/(k\theta))$, where $\alpha$, $\beta$ are constants, $z$ is distance, $k$ is Boltzman's constant and $\theta$ is temperature. The dimension of $\beta$ are

Step 1: Analyze the exponential term.

The argument of an exponential function must be dimensionless.

$$\left[\frac{\alpha z}{k\theta}\right] = M^0 L^0 T^0$$

Step 2: Find the dimensions of the known quantities in the exponent.

The dimension of distance $z$ is length, and the product of Boltzmann's constant $k$ and temperature $\theta$ has the dimensions of energy.

$$[z] = L$$ $$[k\theta] = M L^2 T^{-2}$$

Step 3: Determine the dimensions of $\alpha$.

Solve for the dimensions of $\alpha$ using the dimensionless condition of the exponent.

$$[\alpha] = \frac{[k\theta]}{[z]}$$ $$[\alpha] = \frac{M L^2 T^{-2}}{L}$$ $$[\alpha] = M L T^{-2}$$

Step 4: Determine the dimensions of pressure $P$.

Pressure is defined as force per unit area.

$$[P] = M L^{-1} T^{-2}$$

Step 5: Determine the dimensions of $\beta$.

Since the exponential term is dimensionless, the pre-factor $\frac{\alpha}{\beta}$ must have the same dimensions as pressure $P$.

$$\left[\frac{\alpha}{\beta}\right] = [P]$$ $$[\beta] = \frac{[\alpha]}{[P]}$$ $$[\beta] = \frac{M L T^{-2}}{M L^{-1} T^{-2}}$$ $$[\beta] = L^2$$ $$[\beta] = M^0 L^2 T^0$$

Answer: The dimensions of $\beta$ are $M^0 L^2 T^0$.