An electron possessing a kinetic energy of 100 eV circulates in a path of radius 10 cm within a magnetic field. Determine the magnetic field and the number of revolutions per second completed by the electron.

Step-by-step solution

Kinetic energy of the electron, K = 100 eV = 100 1.6 10⁻¹⁹ J = 1.6 10⁻¹⁷ J The radius of the circular path is given by R = 2mK qB Rearranging for magnetic field B : B = 2mK qR Substituting the standard values ( m = 9.1 10⁻³¹ kg , q = 1.6 10⁻¹⁹ C , R = 0.1 m ): B = 2 9.1 10⁻³¹ 1.6 10⁻¹⁷ 1.6 10⁻¹⁹ 0.1 B = 29.12 10⁻⁴⁸ 1.6 10⁻²⁰ = 5.396 10⁻²⁴ 1.6 10⁻²⁰ 3.37 10⁻⁴ T The number of revolutions per second is the frequency f , given by: f = qB 2 m f = 1.6 10⁻¹⁹ 3.37 10⁻⁴ 2 3.14 9.1 10⁻³¹ f = 5.392 10⁻²³ 57.148 10⁻³¹ 9.43 10^