Question

A neutron beam of kinetic energy 0.04 eV is diffracted at the plane (10 0) of a
crystal having d100 = 3.14 A°. At what glancing angle the 1st order Bragg spectrum
observed ?
(A) 13°11 (8) 14° 9' (C) 13° 45
) None of these​

Answer 1

Answer:

ANSWER IS NONE OF THESE

Answer 2

Given that,

Kinetic energy = 0.04 eV

lattice constant $a= 3.14\ \AA$

Order number = 1

Plane $h,k,l=1,0,0$

We need to calculate the wave length

Using formula of wavelength

$\lambda=\dfrac{h}{\sqrt{2mE}}$

Put the value into the formula

$\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times1.67\times10^{-27}\times0.04\times1.6\times10^{-19}}}$

$\lambda=1.434\times10^{-10}\ m$

We need to calculate the distance

Using formula of distance

$d=\dfrac{a}{\sqrt{h^2+k^2+l^2}}$

Put the value into the formula

$d=\dfrac{3.14\times10^{-10}}{\sqrt{1^2+0+0}}$

$d=3.14\times10^{-10}\ m$

We need to calculate the glancing angle

Using formula of Bragg's law

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

$\sin\theta=\dfrac{n\lambda}{2d}$

Put the value into the formula

$\sin\theta=\dfrac{1\times1.434\times10^{-10}}{2\times3.14\times10^{-10}}$

$\sin\theta=0.2283$

$\theta=\sin^{-1}(0.2283)$

$\theta=13.19^{\circ}$

Hence, The glancing angle is 13.19°.

(D) is correct option