Here is a case where you should take a close look at the magniture of your quantities and do some manipulation to normalize things before throwing the system at the computer:

      Zl = 2.05*10^-15
      α = 1.6381
      ρ = 0.326*10^-10
      k = 8.9875517873681764*10^9
      e = 1.602176565*10^\[Minus]19

divide both of your expressions by `( ρ e )` :  

      f1[x_] = Zl  Exp[-x] / e 
      f2[x_] = α k e/x^2/ρ 

note by the way there is no need for delayed definitions here.
note also I did this by hand, you do not want an expression with ` e^2  /  e `

the expression you are solving is now fairly tame: `f1[x]==f2[x]`

>    12795.1 E^-x == 72.356/x^2

     NSolve[ f1[x] == f2[x], x]

> {{x -> -0.0725216}, {x -> 0.0781981}, {x -> 9.72452}}

with a warning that there could be other solutions, but you can easily convince yourself that there are not.

     Plot[{ f1[x] , f2[x] } , {x, 5, 12}]

![enter image description here][1]

restoring the original definitions we get:

![enter image description here][2]

this result is machine precision 

> 9.724519615727882`

the result I get rationalizing everything (ie. rationalizing each input quantity and never doing any floating operations ) and using `FindRoot[ WorkingPrecision->100 ]` is

> 9.72451961572788227332283528658...

you would be hard pressed to explain how the high precision result is somehow  meaningful.

  [1]: https://i.sstatic.net/k41VS.png
  [2]: https://i.sstatic.net/V3IJR.png