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The defraction of a crystal of Barium with X-ray of wavelength 2.29 \AA gives a first order reflection at 27o8' . What is the distance between the diffracting planes ?

[ sin 27o8'= 0.4561]

  • Option 1)

    1.46 \AA

  • Option 2)

    1.59 \AA

  • Option 3)

    2.51 \AA

  • Option 4)

    5.46 \AA

 

Answers (1)

best_answer

n\lambda =2d\sin \Theta

where n=1\: \: ,\: \: \lambda =2.29A\, ^{\circ} \, \, ,\, \sin 27^{\circ}8=0.4561

putting the value of n,\, \lambda \,  &  \sin \Theta

1\times 2.29=2\times d\times 0.4561

d=\frac{2.29}{2\times 0.4561}=2.51A^{\circ}

 

Bragg's law -

Brag's law is used to find the separation between two layers of a crystal using x-rays.

nλ=2dsinθ

- wherein

Where,

d= distance between two consecutive layer of constituent particles in the crystal

θ= angle of incidence

λ= wavelength of x-ray

n= order of diffraction

 

 


Option 1)

1.46 \AA

This is incorrect

Option 2)

1.59 \AA

This is incorrect

Option 3)

2.51 \AA

This is correct

Option 4)

5.46 \AA

This is incorrect

Posted by

divya.saini

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