In Uranium $\left(Z=92\right)$, the $\mathrm{K}$ absorption edge is $0.107\mathrm{ }\mathrm{\AA }$ and the ${\mathrm{K}}_{\mathrm{\alpha }}$ line is $0.126\mathrm{ }\mathrm{\AA }$. The wavelength of the $\mathrm{L}$ absorption edge is

• Correct answer
  A

  $0.7\mathrm{ }\mathrm{\AA }$
• B

  $1\mathrm{ }\mathrm{\AA }$
• C

  $2\mathrm{ }\mathrm{\AA }$
• D

  $3.2 \mathrm{\AA }$

Solution

Energy of K absorption edge is

$E\left(K\right)=\frac{hc}{{\lambda }_K}=\frac{12.4}{0.107}=115.9 keV$ ,

Energy of Kα line is

$E\left(K_a\right)=E\left(K\right)-E\left(L\right) =\frac{hc}{{\lambda }_a}=\frac{12.4}{0.126}=98.4 keV$

Therefore energy of L absorption edge is

L $E_L=\left(E_K\right)-E\left(K_a\right)=115.9-98.4$

L $E_L=17.5 keV$

The wavelength of the L absorption edge is

${\lambda }_L=\frac{hc}{E_L}=\frac{12.4 keV\AA }{17.5 keV}=0.709 \AA$

Hence, the correct answer is option: 'A'.