Question

In uranium (Z = 92) the K absorption edge is 0.107 Å and the K_a line is 0.126 Å the wavelength of the L absorption edge is-

• A. 0.7 Å
• B. 1 Å
• C. 2 Å
• D. 3.2 Å

Answer 1

Answer: (A)

0.7 Å

Answer 2

E(K) = = $\frac { \mathrm { hc } } { \lambda _ { \mathrm { K } } }$ = $\frac { 12.4 } { 0.107 }$= 115.9 KeV

E(K_a) = E(K) – E(L) = $\frac { \mathrm { hc } } { \lambda _ { \alpha } }$ = $\frac { 12.4 } { 0.126 }$ = 98.4 KeV

E_L = (E_K) – E(K_a)

= 115.9 – 98.4

E_L = 17.5 KeV

l_L = $\frac { \mathrm { hc } } { \mathrm { E } _ { \mathrm { L } } }$ = = 0.709 Å