Question

When X–rays of wavelength $1A^{\circ }$ passes through a gold foil of thickness 2.303 mm, then their intensity reduces to half. The coefficient of absorption for gold(in $m{m}^{-1})$ will be

• A. 0.2
• B. 0.3
• C. 0.4
• D. 0.5

Answer 1

The correct option is B 0.3

Here,$I=\frac{1}{2}{I}_0,$
As, $I=I_0{e}^{-\mu x}$
So,$\mu =\frac{2.303}{x}lo{g}_{10}\frac{I_0}{I}$
$=\frac{2.303}{2.303}\times lo{g}_{10}2=\frac{2.303}{2.303}\times 0.3010=0.30m{m}^{-1}$