The binding energies of K-shell and L-shell electrons in a certain metal are EK and EL, respectively, If a Kαx ray from this metal is incide

Question

The binding energies of K-shell and L-shell electrons in a certain metal are EK and EL, respectively, If a Kαx ray from this metal is incident on a crystal and gives a first-order Bragg reflection at an angle θ measured relative to parallel planes of atoms, what is the spacing between these parallel planes? State your answer in terms of the given variables, using h and c when needed.

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Philomena 3 days 2021-07-20T04:49:25+00:00 1 Answers 2 views 0

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    2021-07-20T04:51:01+00:00

    Answer:

    The separation distance between the parallel planes of an atom is hc/2sinθ(EK – EL)

    Explanation:

    The relationship between energy and wavelength is expressed below:

    E = hc/λ

    λ = hc/EK – EL

    Considering the condition of Bragg’s law:

    2dsinθ = mλ

    For the first order Bragg’s law of reflection:

    2dsinθ = (1)λ

    2dsinθ = hc/EK – EL

    d = hc/2sinθ(EK – EL)

    Where ‘d’ is the separation distance between the parallel planes of an atom, ‘h’ is the Planck’s constant, ‘c’ is the velocity of light, θ is the angle of reflection, ‘EK’ is the energy of the K shell and ‘EL’ is the energy of the K shell.

    Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK – EL)

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