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Question: What is the smallest Bragg angle for x rays of wavelength \(30pm\) to reflect from reflecting planes...

What is the smallest Bragg angle for x rays of wavelength 30pm30pm to reflect from reflecting planes spaced 0.30nm0.30nm apart in a calcite crystal?

Explanation

Solution

Here to solve this problem we have to use the Bragg law where it gives us the ratio between the angle formed due to the ray, distance of the reflecting planes spaced and wavelength of the ray. Now from that we can find the angle formed due to the ray and we need to find the minimum angle where m=1m = 1, it is the whole number of wavelengths which form the consecutive interference.

Complete step by step solution:
As per the problem we have a x rays of wavelength 30pm30pm to reflect from reflecting planes spaced 0.30nm0.30nm apart in a calcite crystal.
Now we need to calculate the smallest Bragg angle.
Now using Bragg’s law we can write,
2dsin⁡θ=mλ2d\sin \theta = m\lambda
Where,
The path difference d is equal to a whole number of m of wavelength due to which the constructive interference will occur. The angle of scattering is equal to θ\theta and the wavelength of the x rays is equal to λ\lambda .
Now from the problem we know,
d=0.30nm=0.30×10−9md = 0.30nm = 0.30 \times {10^{ - 9}}m
λ=30pm=30×10−12m\lambda = 30pm = 30 \times {10^{ - 12}}m
As we need to find the smallest angle θmin⁡{\theta _{\min }}, m=1m = 1
Now putting the known value in the above Bragg’s equation we will get,
2×.30×10−9msin⁡θmin⁡=1×30×10−12m2 \times .30 \times {10^{ - 9}}m\sin {\theta _{\min }} = 1 \times 30 \times {10^{ - 12}}m
Rearranging the above equation we will get,
θmin⁡=sin⁡−11×30×10−12m2×0.30×10−9m{\theta _{\min }} = {\sin ^{ - 1}}\dfrac{{1 \times 30 \times {{10}^{ - 12}}m}}{{2 \times 0.30 \times {{10}^{ - 9}}m}}
On further simplifying we will get,
θmin⁡=sin⁡−150×10−3=sin⁡−10.05{\theta _{\min }} = {\sin ^{ - 1}}50 \times {10^{ - 3}} = {\sin ^{ - 1}}0.05
Hence the smallest Bragg angle θmin⁡=2.9∘{\theta _{\min }} = 2.9^\circ .

Note:
Remember that this law explains the molecular structure and the crystal being identified using the x ray diffraction technique and due to this is gives a relationship between the incident x ray light and its reflection from the crystal surface. Note that this law states that when a x ray light is incident onto a crystal surface then its angle of incidence reflected back with the same angle of scattering.