If the X-rays of wavelength 1.549 $A^0$ will be reflected from crystal having spacing of 4.255 $A^0$, - Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: May 2014

Given:-

$d = 4.255 A^0$

$λ = 1.549 A^0$

To Find:-

Smallest Glancing Angle $(θ_{min})$

Highest Order $(n_{Max})$

Solution:-

Smallest Glancing Angle will be found at n = 1

Using Bragg's Law,

$$ ∴ λ = 2d.sin⁡θ$$

$$∴1.549=2 x 4.255 × sin⁡θ$$

$ ∴ θ_{min} = 10.488^0$

The maximum order of diffraction will occur when sin θ = 1

Therefore, from Bragg's Law, we can write that;

$$n_{Max} λ=2d$$

$$ ∴ n_{Max} = \frac{2d}{λ}$$

$$ ∴ n_{Max} = \frac{2 × 4.255 }{1.549}$$

$$ ∴ n_{Max} = 5.493$$

Hence, we will obtain 5th order maxima and not 6th order maxima

$∴n_{Max} = 5$