Question

It is known that the intensity of x - ray reduces exponentially when passed through any metal medium. If 1 cm of aluminium reduces x - ray intensity by factor of 10, then what thickness in cm is required if incident intensity is doubled to maintain same exit field intensity. [ln 10 = 2.3]

Answer 1

$I=I_0{e}^{-kt}\Rightarrow \frac{I_0}{10}=I_0{e}^{-kt}[t=1cm]$
$e^k=10\Rightarrow ln10=k=2.3$
Again $\frac{I_0}{10}=2I_0{e}^{-k{t}^{\prime }}\Rightarrow e^{kt}=20$

$k{t}^{\prime }$ = ln 20 $\Rightarrow$ 2.3 $t^{\prime }$= ln 2 + ln 10
$2.3t^{\prime }=0.693+2.3\Rightarrow 2.3(t^{\prime }-1)=0.693$

$f^{\prime }=1.30$