Question

A material whose K absorption edge is 0.15 Å is irradiated with 0.1 Å X-rays. The maximum kinetic energy of photoelectrons that are emitted from K-shell is-

• A. 41 KeV
• B. 51 KeV
• C. 61 KeV
• D. 71 KeV

Answer 1

Answer: (A)

41 KeV

Answer 2

|E_K| = = = 82.7 KeV

The energy of incident photon

E_n = = $\frac { 12.4 } { 0.1 }$= 124 KeV

The maximum kinetic energy is

K_max = E_n – |E_K| = 41.3 KeV $\frac { q } { m } = \frac { y v ^ { 2 } } { E L D } = \frac { 2 \times 10 ^ { - 2 } \times \left( 10 ^ { 6 } \right) ^ { 2 } } { 20 \times 10 ^ { 3 } \times 5 \times 10 ^ { - 2 } \times 0.2 } = 10 ^ { 8 } \mathrm { C } / \mathrm { kg }$.41 KeV