Chapter 7, Problem 7.103QE

(a)

Interpretation Introduction

Interpretation:

The velocity of electron required for an electron diffraction experiment if the wavelength is $100\text{ pm}$ has to be determined.

Concept Introduction:

de Broglie established a relation for particles of matter that they also can behave as a wave and has a wavelength. The wavelength of the particle is inversely proportional to its mass. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

  $\text{λ}=\frac{h}{mv}$        (1)

Here,

$\text{λ}$ is a wavelength of particle.

$h$ is a Plank’s constant.

$m$ is a mass of particle.

$v$ is a velocity of mass.

The expression to calculate the root-mean-square speed of particle is as follows:

  $v_{\text{rms}}=\sqrt{\frac{3kT}{m}}$        (2)

Here,

$k$ is a Boltzmann constant and its value is $1.38\times {10}^{-23}{\text{ m}}^2\cdot \text{kg}\cdot {\text{s}}^{-\text{2}}\cdot {\text{K}}^{-\text{1}}$.

$T$ is a temperature.

$m$ is a mass of a particle.

Answer to Problem 7.103QE

The velocity of electron required for an electron diffraction experiment if the wavelength is $100\text{ pm}$ is $7.27\times {10}^6\text{ m/s}$.

Explanation of Solution

Rearrange equation (1) to calculate the value of velocity as follows:

  $v=\frac{h}{m\text{λ}}$        (3)

Substitute $\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}$ for $h$, $9.11\times {10}^{-31}\text{ kg}$ for $m$ and $100\text{ pm}$ for $\text{λ}$ in equation (3).

  $\begin{array}{c}v=\left(\frac{\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}}{\left(9.11\times {10}^{-31}\text{ kg}\right)\left(100\text{ pm}\right)}\right)\left(\frac{1\text{ pm}}{{10}^{-12}\text{ m}}\right)\\ =7.27\times {10}^6\text{ m/s}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The velocity of neutron with wavelength $100\text{ pm}$ has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at $300\text{ K}$ has to be compared.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.103QE

The velocity of neutron with wavelength $100\text{ pm}$ is $3.97\times {10}^3\text{ m/s}$. The root-mean-square velocity at $300\text{ K}$ is $2.73\times {10}^3\text{ m/s}$.

Explanation of Solution

Rearrange equation (1) to calculate value of velocity as follows:

  $v=\frac{h}{m\text{λ}}$        (3)

Substitute $\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}$ for $h$, $1.67\times {10}^{-27}\text{ kg}$ for $m$ and $100\text{ pm}$ for $\text{λ}$ in equation (3).

  $\begin{array}{c}v=\left(\frac{\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}}{\left(1.67\times {10}^{-27}\text{ kg}\right)\left(100\text{ pm}\right)}\right)\left(\frac{1\text{ pm}}{{10}^{-12}\text{ m}}\right)\\ =3.97\times {10}^3\text{ m/s}\end{array}$

Substitute $1.38\times {10}^{-23}{\text{ m}}^2\cdot \text{kg}\cdot {\text{s}}^{-\text{2}}\cdot {\text{K}}^{-\text{1}}$ for $k$, $1.67\times {10}^{-27}\text{ kg}$ for $m$ and $300\text{ K}$ for $T$ in equation (2).

  $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}{\text{ m}}^2\cdot \text{kg}\cdot {\text{s}}^{-\text{2}}\cdot {\text{K}}^{-\text{1}}\right)\left(300\text{ K}\right)}{\left(1.67\times {10}^{-27}\text{ kg}\right)}}\\ =2.73\times {10}^3\text{ m/s}\end{array}$

The value of the velocity of neutron with wavelength $100\text{ pm}$ is higher than the root mean square velocity at $300\text{ K}$.