Why transcendental terms in the laws of nature are dimensionless? Through my years in nuclear engineering, it has always been the case that in physical relations, the arguments of transcendental functions, e.g., the exponential in the law of radioactive decay, $N=N_0 e^{-\lambda t}$, should be dimensionless so that the outcome is correspondingly dimensionless.
In purely empirical relations, however, I have observed that it's okay for transcendental functions to have dimensional arguments as long as they fit the results of experiments. For example, the energy spectrum of prompt fission neutrons is fitted by:
$$\chi(E)=0.453 e^{-1.036E}\text{sinh}\sqrt{2.29E}.$$
Or does the 1.036 in the exponential have units of ${\text{eV}}^{-1}$? 
Is my observation correct? And if so, what is the basis of all that? I tried to dig into dimensional analysis, but I couldn't understand much.
Update on Jan 14, 2020:
I've found an insight on Wikipedia that I think will, hopefully, enrich the discussion.
A problem with transcendental functions is that: applying a non-algebraic operation to a dimensional quantity creates paradoxical results. For instance, $\text{log}_a(5L)=\text{log}_a(3)+\text{log}_a(L)$, where $L$ is the dimension of length, and $a$ is an arbitrary base.
This also raises the question: is $\text{log}_a(L)$ dimensional? Does a transcendental function become dimensional when fed with a dimensional argument?
 A: For a theory to make sense, it has to make useful predictions regardless of the units you use. 
For algebraic formulae, that usually ends giving you constants in the formula with different values. Newton’s gravity looks exactly the same in English and metric units, except the constant G has a different numeric value. 
But that doesn’t work for transcendental functions. $e^{24}$ when working in inches and $e^{2}$ when working in feet can’t be made equal with a constant in front of the expression; that number would be different for different input values, hence not constant. 
Instead, we specify what units the argument should be. “Must be in feet”, if that makes the numbers work, provides a convention. 
Sometimes we’ll actually write that as 
$$e^{x/(\rm{1 ft})}$$
Then if x is in other units, you’re reminded to convert.
Sometimes, this happens automatically. For example a lifetime:
$$e^{-t/\tau}$$
Both of those quantities have units, and we’re used to making them consistent before dividing. Once we do that, it becomes “3 lifetimes” and the units we measure in no longer matter: the transcendental argument is again unitless. 
A: You are correct,  in the case you cite, 1.036 has dimension eV${}^{-1}$ (assuming $E$ is measured in eV).
All functions can be written in terms of dimensionless arguments, not just the transcendental ones.  It's common to write equations in dimensionless form, where all variables are dimensionless, that is, the ratio of some measured quantity to a "characteristic value" of the same dimension that sets the scale.  
In your case the characteristic value is $E_0 = (1/1.036) eV$, and it sets the scale.   Values of $E$ less than $E_0$ are "small", those greater are "large".  When $E=E_0$ the value of the function is of the order of unity (one).
Every function can be written with dimensionless arguments.  For "normal" rational, power law, algebraic, etc functions, the characteristic value can be factored out if desired, and the dimensionless quality is less evident.
A: $$e^x = 1 + x + x^2 + ...$$
Therefore if $x$ has units, $e^x$ doesn't even make sense by dimensional analysis. The argument of log, and any trigonometric function is also unitless for the same reason.
