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3 years ago

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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 met

al 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.

1 answer:

Svetach [21]3 years ago

5 0

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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Answer:

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Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

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$=> \ \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}$

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3 years ago

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