Prove that fur any sine function sin(kx+ϕ)of wavelength shorter than 2a, where ais the atomic spacing. there is a sine function with a wavelength longer than 2a that has the same values at the points x = a , 2a , 3a . and so on. (Note: It is probably easier to work with wave number than with wavelength. We sick to show that for every wave number greater than there is an equivalent less than π/a.)

Short Answer

Expert verified

It is proved that there is a sine function with a wavelength less than 2a.

Step by step solution

01

Sum of angle and expression for θ:

The atomic spacing is a.

The expression of sum of angles is given by,

θ1+θ2=nπ

The expression for is given by,

θ=kx+ϕ

02

Determine sum of angles:

The expression of sum of angles is calculated as,

θ1+θ2=nπnk1x+ϕ+k2x+ϕ=nπnk1x+k2x+2ϕ=nπn..........(1)

For

πana+ϕ=nπ2ϕ=nπ2-nπ=-nπ2

03

solve further:

Substitute -nπ2for ϕin equation (1).

k1+k2x+2-nπ2=nπk1+k2na=2k1+k2=2πak1=2πa-k2

Now since,

k1>πa,2πa-k2>πa,k2<πa

Therefore, it is proved that there is sine function with wavelength less than 2a.

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