When 0.50 Ã X-rays strike a material, the photoelectrons from the k shell are observed to move in a circle of radius 23 ~mm in a magnetic field of 2 10⁻² tesla acting perpendicularly to the direction of emission of photoelectrons. What is the binding energy of k-shell electrons?

Step-by-step solution

As we know, F = qvB = m v ² R v = q m BR The kinetic energy of the photoelectron array l = 1 2 m v²= 1 2 e² B² R² m = 1 2 (1.6 10⁻¹⁹ )² (2 10⁻² )² (23 10⁻³ )² (9.1 10⁻³¹ ) =2.97 10⁻¹⁵ ~J = 2.97 10⁻¹⁵ 1.6 10⁻¹⁹ =18.36 keV array Fnergy of the incident n = hc = 12.4 0.50 =24.8 keV Therefore, Binding energy =24.8-18.6=6.2 keV