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 $56.4^{\circ}$

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.