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. 7.4keV
• D. 8.6 keV

Answer 1

Answer: (A)

6.2 keV

Answer 2

evB = ̃ v = $\frac { \mathrm { e } } { \mathrm { m } }$BR

K.E. of photoelectrons

K= $\frac { 1 } { 2 }$mv² = = 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 KeV