Question

Calculate the velocity and kinetic energy with which the electrons strike the target of an x-ray tube operated at 50,000 volts. What is the short-wavelength limit of the continuous spectrum emitted and the maximum energy per quantum of radiation?

Answer 1

Given potential difference, V = 50000 Volt

Charge of electron, q = e = 1.602x10^-19 C

Kinetic energy and potential difference V are related as

KE = q x V = e x V = 1.602x10^-19 C x 50000 Volt = 8.01x10^-15 J or 50000 eV (answer)

mass of electron, m = 9.11 x 10^-31 Kg

Also KE = 1/2 x mv² = 8.01x10^-15 J

=> 1/2 x 9.11 x 10^-31 Kgxv² = 8.01x10^-15 J

=> v = underroot (2x8.01x10^-15 J /  9.11 x 10^-31 Kg )

=> v = 1.33 x 10⁸ m/s (answer)

The shortest wavelength can be calculated as

KE = hc /

=> = hc / KE =  6.626 x10^-34 J.sx 3x10⁸ m/s / 8.01x10^-15 J = 2.48 x 10^-11 m