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.

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)

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft