Question

When 24.8 KeV x-rays strike a material, the photoelectrons emitted from K shell are observed to move in a circle of radius 23 mm in a magnetic field of 2 × 10^–2 T. The binding energy of K-shell electrons is-

• A. 6.2 KeV
• B. 5.4 KeV
• C. 7.4 KeV
• D. 8.6 KeV

Answer 1

Answer: (A)

6.2 KeV

Answer 2

evB = $\frac{mv^{2}}{R}$̃ v = $\frac{e}{m}BR$

K.E. of photoelectrons

K = $\frac{1}{2}mv^{2}$ = $\frac{e^{2}B^{2}R^{2}}{2m}$= 2.97 × 10^–15 J

K.E. = 2.97 × 10^–15 J = 18.6 KeV

K.E. = E_P – E(K)

E(K) = E_P – K.E. = 24.8 – 18.6 = 6.2 K.E.