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25 October, 09:09

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.)

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  1. 25 October, 11:24
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    Answer: 4.0024 x 10^ - 11 m or 0.040024 nm

    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
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