What do you mean by "normalized"
Your factor is
In[76]:= nn = 1/(12. 3.14 10^-8)
Out[76]= 2.65393*10^6
with this I get
In[80]:= Exp[-(((x - 5)^2 - 1)/nn)] /. x -> 0
Exp[-(((x - 5)^2 - 1)/nn)] /. x -> 25
Out[80]= 0.999991
Out[81]= 0.99985
Seems your expression is virtually constant, so, no contours