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  • A superconductor has a lattice spacing of . What is the angle of incidence (in degrees) for X-rays of w

A superconductor has a lattice spacing of 3.0\; A. What is the angle of incidence (in degrees) for X-rays of wavelength 1.5\; A. that undergo a first-order diffraction from the crystal surface, according to Bragg's equation?

 

Option: 1

25.2^{\circ}


Option: 2

31.4^{\circ}


Option: 3

42.0^{\circ}


Option: 4

56.4^{\circ}


Answers (1)

best_answer

Bragg's equation relates the wavelength of X-rays to the angle of incidence and the lattice spacing of a crystal:

2 \: \: d \sin \theta=n \lambda

where d is the lattice spacing, \theta  is the angle of incidence, n is the order of diffraction, and \lambda is the wavelength of the incident X-rays.

For a first-order diffraction (n = 1) from a superconductor with a lattice spacing of 3.0\: A and X-rays of wavelength 1.5\: \: A,Bragg's equation becomes:

2(3.0 \AA) \sin \theta=(1)(1.5 \AA)

Simplifying and solving for \theta,we get

\sin \theta=\frac{1}{4} \theta=\sin ^{-1} \frac{1}{4} \theta \approx 31.4^{\circ}

Hence, option b) 31.4° is the correct answer.

Posted by

Rakesh

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