Determine the interplanar spacing of the first peak, and the atomic radius of th

Question

Determine the interplanar spacing of the first peak, and the atomic radius of the element

Visible light passes through a hole of diameter 150 nm in a sheet of metal and falls on a distant screen. If the metal is heated, making the hole larger, the size of the central bright spot on the screen gets smaller. True False I When a beam of light, which is traveling in glass, strikes an air boundary, there is a 45degree phase change in the reflected beam. a 90degree phase change in the reflected beam. a -45degree phase change in the reflected beam. no phase change in the reflected beam. 180degree phase change in the reflected beam.

Explanation / Answer

The first peak seems to be at 2 = 430.

So, = 43/2 = 21.50

Braggs law; 2dSin = n* lambda ( lamba is the wavelength of the light, 0.154 nm)

Here n=1 , first order reflection

So, d = 0.154 / 2*Sin = 0.154 / 2*0.366 = 0.154/ 0.732 = 0.210 nm.

Interplanar spacing of the first peak is 0.210 nm.

To determine the atomic radius of the element we need to determine a (lattice constant) and know whether it is Primitive or FCC or BCC.

As we have determined the d for first peak we can determine 'd' for all the other peaks and hence the 'd2' for all the peaks.

After determining all the d2, we need to multiply it with some constant ( different constants for different d2 s) so that the d2 s when multiplied with the different constants gives almost the same value. The constant will give the value of h, k, l for the different planes ( follow h2 + k2 + l2 = constant, for a particular plane).

The sum of h,k,l is even: BCC

h,k,l : all even or all odd --> FCC

The multiplication of d2 s with the different constants will give the value of almost same a2s and take the average of a2s and the root of the average of a2 s will give the lattice constant (a).

As determined by the values of h,k,l we know the Crystal lattice and using the relation of the lattice constant and the radius of the element, we can determine the radius of the element.

Thank You

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