Answer:
The  value is  [tex]\alpha =  0.002 Np/m[/tex]
Explanation:
From the question we are told that
 The first amplitude of the wave is  [tex]E_{max}1 =  98.02 \  V/m[/tex]
 The first  depth  is  [tex]D_1 =  10 \  m[/tex]
  The second amplitude is  [tex]E_{max}2 =  81.87 \  (V/m)[/tex]
  The second depth is [tex]D_2 = 100 \ m[/tex]
Generally from the spatial wave equation we have
  [tex]v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)[/tex]
=>    [tex]\frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}[/tex]
So considering the ratio of the equation for the  two depth
[tex]\frac{A}{A_S} Â = Â \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}[/tex]
=> Â [tex]\frac{98.02}{81.87} Â = Â \frac{e^{-10 \alpha }}{e^{-100 \alpha }}[/tex]
=>  [tex]\alpha  =  \frac{0.18}{90}[/tex]
=> Â Â [tex]\alpha = Â 0.002 Np/m[/tex]
   Â
