If $E$ and $E_0$ denote energies at time $t$ and $t_0$ respectively, and $L$ and $L_0$ distance from some point at $t$ and $t_0$ respectively, then which of the following equations can be declared to be incorrect on dimensional grounds

(A) $E=\frac{2E_{0}L}{L_0}$

(B) $E=E_0{e}^{-\frac{2L}{L_0}}$

(C) $E=2L{e}^{\frac{-L}{E_0}}$

(D) $E=2\left(\frac{E_0}{L_0}\right)\times e^{\frac{-L}{L_0}}$

Dimensions of energy $=\left[E\right]=\left[E_0\right]=\left[M{L}^2{T}^{-2}\right]$

Dimensions of length

$=\left[L\right]=\left[L_0\right]=\left[M^0{L}^1{T}^0\right]$

For A:

LHS $=\left[M{L}^2{T}^{-2}\right]$

RHS $=\frac{\left[M{L}^2{T}^{-2}\right]\left[L\right]}{\left[L\right]}=\left[M{L}^2{T}^{-2}\right]$

In this case, LHS=RHS

For B:

Exponent should be dimensionless. As we can see

$\left[\frac{-2L}{L_0}\right]=\left[\frac{L}{L}\right]=\left[M^0{L}^0{T}^0\right]$

and LHS $=\left[M{L}^2{T}^{-2}\right]$

and RHS $=\left[M{L}^2{T}^{-2}\right]$

So in this case, LHS=RHS

For C:

Exponent should be dimensionless but here

$\left[\frac{-L_0}{E_0}\right]=\frac{\left[M^0{L}^1{T}^0\right]}{\left[M{L}^2{T}^{-2}\right]}=\left[M^{-1}{L}^{-1}{T}^2\right]$

is not dimensionless therefore it is wrong.

For D:

Exponent should be dimensionless. Here

$\left[\frac{-L}{L_0}\right]=\frac{\left[M^0{L}^0{T}^0\right]}{\left[M^0{L}^0{T}^0\right]}=\left[M^0{L}^0{T}^0\right]$

Now, LHS $=\left[E\right]=\left[M{L}^2{T}^{-2}\right]$

and RHS $=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L\right]}=\left[M{L}^1{T}^{-2}\right]$

As in this case, LHS $\ne$RHS, hence it is wrong.