In an x-ray tube with 70kv, what is the velocity of the electrons upon reaching the anode? and what is the shortest wavelength of the x-rays emitted from this tube?

v=3.92*10¹⁷ m/s, and l=2.93*10⁻³⁰m

Explanation:

The velocity of the electrons in the x-ray tube is related to the mass of electrons and the energy applied, the equation is E=0.5*m*v², where E is the energy in kV, m is the mass of the electrons (9.11x10-31kg), and v the velocity. Substituting, 70000 = 0.5*(9.11x10-31kg)*v², and rearranging the terms, sqrt((70000)/(0.5)(9.11x10-31)) = v = 3.92*10¹⁷ m/s. Then the energy is related to the constant h (6.62607015×10−34 Js), the speed of light 299792458 m/s, and "l" that is the wavelength. So the equation is l=(h*c)/E, and l = 2.93*10⁻³⁰m.

i believe it is c.) 1575 j.