Chapter 28, Problem 78P

(a)

To determine

The wavelength of the neutron that undergoes strong Bragg’s diffraction.

Answer to Problem 78P

The wavelengths of the neutron are $6.9\times {10}^{-11}\text{m, 3}\text{.5}\times {10}^{-11}\text{m}\text{and 2}\text{.3}\times {10}^{-11}\text{m}$.

Explanation of Solution

The speed range of the neutron is $0\text{ to }2.0\times {10}^4{\text{ms}}^{-1}$, the interplanar distance of the crystal is $0.20\text{nm}$ and the angle of diffraction is ${10.0}^{\circ }$.

Write the expression for de Broglie wavelength

$\lambda =\frac{h}{mv}$                                           (I)

Here, $\lambda$ is the de Broglie wavelength, $h$ is the Planck’s constant, $m$ is the mass of the neutron and $v$ is the velocity.

The de Broglie wavelength is inversely proportional to velocity and hence greater velocity corresponds to smaller wavelength and vice versa.

Substitute $1.675\times {10}^{-27}\text{kg}$ for $m$, $2.0\times {10}^4{\text{ms}}^{-1}$ for $v$ and $6.626\times {10}^{-34}\text{ Js}$ for $h$ in (II) to find $\lambda$

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{ Js}}{1.675\times {10}^{-27}\text{kg}\times 2.0\times {10}^4{\text{ms}}^{-1}}\\ =\frac{6.626\times {10}^{-34}{\text{ kgm}}^2{\text{s}}^{-1}}{1.675\times {10}^{-27}\text{kg}\times 2.0\times {10}^4{\text{ms}}^{-1}}\\ =2.0\times {10}^{-11}\text{ m}\end{array}$

Thus, the minimum de Broglie wavelength of the neutron is $2.0\times {10}^{-11}\text{ m}$.

Substitute $1.675\times {10}^{-27}\text{kg}$ for $m$, $0$ for $v$  and $6.626\times {10}^{-34}\text{ Js}$ for $h$ in (II) to find $\lambda$

$\lambda =\frac{6.626\times {10}^{-34}\text{ Js}}{1.675\times {10}^{-27}\text{kg}\times 0{\text{ms}}^{-1}}$

Thus, the maximum de Broglie wavelength of the neutron is $\text{infinite}$.

Thus, the range of de Broglie wavelength lies between $2.0\times {10}^{-11}\text{ m to }\infty$

Write Bragg’s law to find wavelength

$n\lambda =2d\sin \theta$                                          (II)

Here, $n$ is the order of diffraction, $d$ is the interplanar distance and $\theta$ is the angle of diffraction.

Rearranging (II)

$\lambda =\frac{2d\sin \theta }{n}$                                         (III)

Substituting $0.20\text{nm}$ for $d$ and ${10.0}^{\circ }$ for $\theta$ and $1$ for $n$ in (III) to find $\lambda$

$\begin{array}{c}\lambda =\frac{2\times 0.20\text{nm}\times \sin {10.0}^{\circ }}{1}\\ =6.9\times {10}^{-11}\text{m}\end{array}$

Substituting $0.20\text{nm}$ for $d$ and ${10.0}^{\circ }$ for $\theta$ and $2$ for $n$ in (III) to find $\lambda$

$\begin{array}{c}\lambda =\frac{2\times 0.20\text{nm}\times \sin {10.0}^{\circ }}{2}\\ =3.5\times {10}^{-11}\text{m}\end{array}$

Substituting $0.20\text{nm}$ for $d$ and ${10.0}^{\circ }$ for $\theta$ and $3$ for $n$ in (III) to find $\lambda$

$\begin{array}{c}\lambda =\frac{2\times 0.20\text{nm}\times \sin {10.0}^{\circ }}{3}\\ =2.3\times {10}^{-11}\text{m}\end{array}$

Substituting $0.20\text{nm}$ for $d$ and ${10.0}^{\circ }$ for $\theta$ and $4$ for $n$ in (III) to find $\lambda$

$\begin{array}{c}\lambda =\frac{2\times 0.20\text{nm}\times \sin {10.0}^{\circ }}{4}\\ =1.7\times {10}^{-11}\text{m}\end{array}$

From the above calculated wavelengths, the wavelengths fitting the range of de Broglie wavelength are only the first three.

Thus, the wavelengths of the neutron that undergoes strong Bragg’s diffraction are $6.9\times {10}^{-11}\text{m, 3}\text{.5}\times {10}^{-11}\text{m}\text{and 2}\text{.3}\times {10}^{-11}\text{m}$.

(b)

To determine

The angle of emergence with respect to the incident beam.

Answer to Problem 78P

The angle of emergence is ${20.0}^{\circ }$.

Explanation of Solution

The angle of incidence is ${10.0}^{\circ }$.

Angle of reflection tells that the angle of incidence and angle of reflection is same for plane surfaces.

The vertically opposite angles are equal and hence the angle of emergence with respect to the incident beam is twice the angle of incidence i.e. $2\left({10.0}^{\circ }\right)$.

Thus the angle of emergence is ${20.0}^{\circ }$ for all the three wavelengths.