99 Percentile Qs Bank for JEE MainPhysicsNuclear Physics
When 24.8 k e V 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
Options
- A6.2 k e V
- B5.4 k e V
- C7.4 k e V
- D8.6 k e V
Correct answer
A. 6.2 k e V
Step-by-step solution
evB = mv 2 R ⇒ v = e m BR Kinetic energy of photoelectrons K = 1 2 mv 2 = e 2 B 2 R 2 2 m K = 2 .97 × 10 - 15 J = 18 .6 KeV K = E P - E K E K = E P - K = 24 .8 - 18 .6 = 6 .2 KeV