Question

An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field towards the other end. If they move fast enough when they strike the positive electrode at the other end, they will give up their energy as X-Rays
(a) Through what potential difference should electrons be accelerated so that their speed is 1% of the speed of light?
(b) What potential difference would be needed to give the protons same kinetic energy as electrons?
(c) What speed would this potential difference give to the protons, both in m/s and as a % of the speed of light.

Answer 1

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = $\frac{1}{2}$ mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = $\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }$

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             $K_e = K_p$

              K_p = ½ m v_e²

              K_p = $\frac{1}{2}$  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = $\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }$

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = $\frac{7 \ 10^4 }{ 3 \ 10^8}$

        v/c= 2.33 10⁻⁴