Chapter 28, Problem 87P

(a)

To determine

The de Broglie wavelength of the electron.

Answer to Problem 87P

The de Broglie wavelength of the electron is $167\text{}\text{ pm}$.

Explanation of Solution

The accelerating potential is $\text{54}\text{.0 V}$.

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 electron and $v$ is its velocity.

Write the expression for kinetic energy

$K=\frac{1}{2}m{v}^2$                                           (II)

Here, $K$ is the kinetic energy of the electron.

Rearranging the above equation

$v=\sqrt{\frac{2K}{m}}$                                           (III)

Substituting (III) in (I)

$\lambda =\frac{h}{m}\sqrt{\frac{m}{2K}}$

$\lambda =\frac{h}{\sqrt{2mK}}$                                           (IV)

Write the expression for kinetic energy in terms of accelerating potential

$K=qV$                                           (V)

Here $q$ is the charge and $V$ is the accelerating potential difference.

Substituting (V) in (VI)

$\lambda =\frac{h}{\sqrt{2mqV}}$                                           (VI)

Substitute $9.109\times {10}^{-31}\text{ kg}$ for $m$, $1.602\times {10}^{-19}\text{ C}$ for $q$, $\text{54}\text{.0 V}$ for $V$,  and $6.626\times {10}^{-34}\text{ Js}$ for $h$ in (VI) to find $\lambda$

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{ Js}}{\sqrt{2\times 9.109\times {10}^{-31}\text{ kg}\times 1.602\times {10}^{-19}\text{ C}\times \text{54}\text{.0 V}}}\\ =167\times {10}^{-12}\text{m}\\ =167\text{}\text{ pm}\end{array}$

Thus, the de Broglie wavelength of the electron is $167\text{}\text{ pm}$.

(b)

To determine

Find the diffraction angle for first order maximum in Bragg’s diffraction.

Answer to Problem 87P

The Bragg angle for first order maximum is ${66.5}^{\circ }$.

Explanation of Solution

Write Bragg’s law to find Bragg angle

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

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

Rearranging (VII) for $\theta$

$\theta ={\sin }^{-1}\left(\frac{n\lambda }{2d}\right)$                                         (VIII)

Substitute $1$ for $n$, $167\text{}\text{ pm}$ for $\lambda$ and $0.091\text{ nm}$ for $d$ in (VIII) to find $\theta$

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1\times 167\text{}\text{ pm}}{2\times 0.091\text{ nm}}\right)\\ ={\sin }^{-1}\left(\frac{167\times {10}^{-12}\text{}\text{ m}}{2\times 0.091\times {10}^{-9}\text{ m}}\right)\\ ={66.5}^{\circ }\end{array}$

Thus, the Bragg angle is ${66.5}^{\circ }$.

(c)

To determine

Check whether the scattering angle is ${130}^{\circ }$.

Answer to Problem 87P

Yes. The Bragg’s scattering angle agrees with the observed scattering angle.

Explanation of Solution

The Bragg angle is ${66.5}^{\circ }$ and the scattering angle is ${133}^{\circ }$.

A sketch of the Bragg diffraction showing the relation between the Bragg angle and the scattering angle.

Write the phase difference between the incident beam and the scattered beam (refer figure 1)

${\varphi }_a=2\theta$                                         (IX)

Here, ${\varphi }_a$ is thee scattering angle calculated using figure 1

Substituting ${66.5}^{\circ }$ for $\theta$ in (IX) to find ${\varphi }_a$ 

$\begin{array}{c}{\varphi }_a=2\left({66.5}^{\circ }\right)\\ ={130}^{\circ }\end{array}$

Write the expression for percentage error

$\%\text{ error of }\varphi =\frac{\left|{\varphi }_a-\varphi \right|}{\varphi }\times 100\%$                                         (X)

Here, ${\varphi }_a$ is the approximate value of scattering angle and $\varphi$ is the actual scattering angle

Substituting ${130}^{\circ }$ for ${\varphi }_a$  and ${133}^{\circ }$ for $\varphi$ in (X)

$\begin{array}{c}\%\text{ error of }\varphi =\frac{\left|{130}^{\circ }-{133}^{\circ }\right|}{{133}^{\circ }}\times 100\%\\ =2.3\%\end{array}$

Since the percentage error is small, the Bragg’s scattering angle agrees with the actual scattering angle.