Use the fact that efficiency is dimensionless

Since Carnot efficiency $\eta$ is a pure number, $[\eta]=1$ Also, $[\sin\theta]=1, \qquad \left[\log_e(\cdot)\right]=1$ because both sine and logarithm require dimensionless arguments and give dimensionless values.

Make the logarithm argument dimensionless

Given $\log_e\left(\frac{\beta x}{kT}\right)$ its argument must be dimensionless: $\left[\frac{\beta x}{kT}\right]=1$ Hence, $[\beta x]=[kT]$

Now,

• $[x]=L$
• Boltzmann constant $k$ has dimensions of energy per temperature, so $[kT]=\text{energy}=ML^2T^{-2}$

Therefore, $[\beta]=\frac{[kT]}{[x]}=\frac{ML^2T^{-2}}{L}=MLT^{-2}$

So $\beta$ has the dimensions of force.

Hence Option A is correct.

Use overall dimensionlessness of $\eta$

From $\eta=\frac{\alpha\beta}{\sin\theta}\log_e\left(\frac{\beta x}{kT}\right)$ and since $\sin\theta$ and the logarithm are dimensionless, $[\alpha\beta]=1$ Therefore, $[\alpha]=[\beta]^{-1}=(MLT^{-2})^{-1}=M^{-1}L^{-1}T^2$

Check Option B

We need dimensions of $\alpha^{-1}x$.

Since $[\alpha^{-1}]=[\beta]=MLT^{-2}$ then $[\alpha^{-1}x]=(MLT^{-2})(L)=ML^2T^{-2}$ which is the dimension of energy.

Hence Option B is correct.

Check Option C

Since $\eta$ and $\sin\theta$ are both dimensionless, $[\eta^{-1}\sin\theta]=1$ Also from Step 3, $[\alpha\beta]=1$ So both have the same dimensions.

Hence Option C is correct.

Check Option D

We found: $[\beta]=MLT^{-2}$ but $[\alpha]=M^{-1}L^{-1}T^2$ These are reciprocals, not the same.

Hence Option D is incorrect.

Final conclusion

The incorrect option is: $\boxed{D}$