Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Trending News

Promoted
Jono
Jono asked in Science & MathematicsPhysics · 7 years ago

How do i do this wavelength question?

There are electron sources of 10kV, 20kV and 30kV. What is the minimum wavelength radiation emitted by the 20kV source.

Can someone please show me how to work this out? thanks.

1 Answer

Relevance
  • peterpan
    Lv 7
    7 years ago

    E=p/c=mv/c

    E=hf=hv/λ

    then

    h/λ=m/c

    λ=hc/m

    where

    m=m0/(1-v^2/c^2)^1/2

    v^2=2eV/m

    then

    m=(m0+4e^2V^2/c^2)^1/2

    V=20 kV

    m0=9.11x10^-31 kg

    e=1.6x10^-19 C

    h=6.626x10^-34 Js [Planck constant]

    c=3x10^8 m/s [speed of light]

    put in the numbers and you get the result

Still have questions? Get your answers by asking now.