Question

Bragg reflection results in a first-order maximum at 15.0°. In this case, at what angle would the second-order maximum occur?

Answer 1

31.174°

Step-by-step explanation:

Bragg's condition is occur when the wavelength of radiation is comparable with the atomic spacing.

So, Bragg's reflection condition for n order is,

`2dsin\theta=n\lambda`

Here, n is the order of maxima,
`\lambda` is the wavelength of incident radiation, d is the inter planar spacing.

Now according to the question, first order maxima occur at angle of 15°.

Therefore

`2dsin(15^(\circ))=\lambda\\sin(15^(\circ))=(\lambda)/(2d)`

Now for second order maxima, n=2.

`2dsin\theta=2* \lambda\\sin\theta=(2\lambda)/(2d)`

Put the values from above conditions

`sin\theta=2* sin(15^(\circ))\\sin\theta=2* 0.258819045103\\sin\theta=0.517638090206\\\theta=sin^(-1)0.517638090206\\ \theta=31.174^(\circ)`

Therefore, the second order maxima occurs at 31.174° angle.