Question

A color television tube generates some X-rays when its electron beam strikes the screen. What is the shortest wavelength of these X-rays, in meters, if a 31 kV potential is used to accelerate the electrons? (Note that TVs have shielding to prevent these X-rays from reaching viewers.)

Answer 1

Answer: 4.0024 x 10^ -11 m or 0.040024 nm

Step-by-step explanation:

λ = h c/ΔE

λ = wave lenght

h = 6.626 x 10 ^ -34 m² kg /s = planck constant

ΔE = 31 keV potential ( 1 keV = 1.6021 x 10^-16J)

c = velocity of light = 3 x 10⁸ m/s

substitute gives

λ = 6.626 x 10 ^ -34 m² kg /s x 3 x 10⁸ m/s = 4.0024 x 10^ -11 m

31 x 1.6021x10^-16 J