Calculation of true thickness of the bed

49) In a cratonic region, radioactive heat generation decreases exponentially with depth. Assuming characteristic depth as 10km and surface heat generation as $3\mu Wm^{-3}$ and neglecting mantle heat flow , the heat production per unit volume for a 30 km thick  layer will be_______________$\mu Wm^{-3}$ .[round off to 2 decimal places].

(Thanks to Chandrasekhar and Rajkumar, AKNU)

 Solution:

 Given that Depth(D)= 10 km

  Layer thickness(Z) = 30 km

 we know that

$A(z)=A_{0}e^{\frac{-Z}{D}}$

by integrating both sides of the above equation with repspect to the thickness layer from 0 to 30 km then we get

$\int_{0}^{30}A(z) z =A_{0}\int_{0}^{30}e^{\frac{-Z}{D}}dz.$

 

$\int_{0}^{30}A(z) z =A_{0}(\frac{e^{\frac{-Z}{D}}}{-1/D})$

 

$[A(z) z]_0^{30} =-A_{0}D[e^\frac{-Z}{D}]_{0}^{30}$

 

$A(z) [30-0] = -A_{0}D [e^\frac{-30}{10}-e^\frac{0}{10}]$

 

$30 A(z)   = -A_{0}D[e^{-3}-1]$

 

$30A(z)   =A_{0}D[1-e^{-3}]$

 

$30A(z)   =A_{0}D[0.95]$

 

   $A(z) =\frac{A_{0}D}{30}[0.95]$

  

   $A(z)=\frac{3\times10}{30}\times0.95$

  

$A(z) =0.95  \mu Wm^{-3}$