Chapter 7, Problem 7.2P

(a)

Interpretation Introduction

Interpretation:

The de Broglie wavelength for electron accelerated through $1V$ and also the momentum for the electrons has to be calculated.

Concept Introduction:

Work function: Work function of a metal is measure of strength of electrons is held in a metal atom.

$\text{hν}\text{=}\text{KE}\text{+}\varphi \boxed{\begin{array}{c}here,\varphi =\text{ work function}\\ \text{KE }=\text{ kinetic energy}\\ \text{h = Planck’s constant} \\ \text{ν}=\text{ frequency}\end{array}}$

Energy can be in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.

${\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\boxed{\begin{array}{l}here,\text{m - Mass in kilograms}\hfill \\ \text{v - Velocity in meters per second}\hfill \end{array}}$

Louis de Broglie in $1923$ rationalized that when light shows particle aspects, then particles of matter display properties of waves under definite circumstances.

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\\ \boxed{\begin{array}{l}\text{h is Planck’s constant }\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\text{ which relates energy and frequency}\\ \text{υ}\text{is the speed of particle}\\ \text{m is the mass of particle }\\ \text{λ}\text{is the wavelength}\\ \text{p is the momentum}\end{array}}\end{array}$

The above equation is called de Broglie relation.

Explanation of Solution

Given:

$10MeV$ Particle accelerator

The electron wavelength present in given accelerator is determined by using kinetic energy equation.

$\begin{array}{c}{\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{m}}_{\text{e}}{\text{v}}^{\text{2}}\\ \text{p}={\text{m}}_{\text{e}}\text{v}\\ {\text{E}}_{\text{k}}\text{=}\frac{{\text{p}}^{\text{2}}}{{\text{2m}}_{\text{e}}}\\ \text{p}=\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}\end{array}$

$\text{p}=\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}\to 1$

Substituting equation 1 in de Broglie equation

$\begin{array}{c}\text{p}=\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}\to 1\\ \text{λ=}\frac{\text{h}}{\text{p}}\Rightarrow \text{de}\text{Broglie}\text{equation}\\ =\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}}\to 2\end{array}$

For accelerated electron energy is equal to $e\Delta \varphi$. Then equation 2 becomes,

$\begin{array}{c}=\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}}\to 2\\ =\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}\text{eΔ}\varphi }}\left[{\text{E}}_{\text{k}}=\text{eΔ}\varphi \right]\\ \text{λ}=\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}\text{eΔ}\varphi }}\end{array}$

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}\text{eΔ}\varphi }}\\ =\frac{\text{6}\text{.626}\times {\text{10}}^{\text{-34}}}{\sqrt{\text{2}\left(9.109\times {10}^{-31}\right)\left(1.602\times {10}^{-19}\right)\left(1\right)}}\times \frac{\text{J}\text{s}}{\sqrt{\text{kg}\text{C}\text{eV}}}\left[\text{1J=}\text{1}\text{CV=1}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\right]\\ =1.23\times {10}^{-9}\text{m}\\ \text{=}1.23\text{nm}\left[\text{1nm}\text{=}{\text{10}}^{\text{-9}}\text{m}\right]\\ \end{array}$

$\begin{array}{c}\text{p}=\frac{\text{h}}{\text{λ}}\\ =\frac{\text{6}\text{.626}\times {\text{10}}^{\text{-34}}}{1.23\times {10}^{-9}}\times \frac{\text{J}\text{s}}{\text{m}}\left[\text{1J=}\text{1}\text{CV=1}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\right]\\ =5.4\times {10}^{-25}\text{kg}\text{m}{\text{s}}^{\text{-1}}\\ \end{array}$

(b)

Interpretation Introduction

Interpretation:

The de Broglie wavelength for electron accelerated through $1kV$ and also the momentum for the electrons has to be calculated.

Concept Introduction:

Work function: Work function of a metal is measure of strength of electrons is held in a metal atom.

$\text{hν}\text{=}\text{KE}\text{+}\varphi \boxed{\begin{array}{c}here,\varphi =\text{ work function}\\ \text{KE }=\text{ kinetic energy}\\ \text{h = Planck’s constant} \\ \text{ν}=\text{ frequency}\end{array}}$

Energy can be in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.

${\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\boxed{\begin{array}{l}here,\text{m - Mass in kilograms}\hfill \\ \text{v - Velocity in meters per second}\hfill \end{array}}$

Louis de Broglie in $1923$ rationalized that when light shows particle aspects, then particles of matter display properties of waves under definite circumstances.

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\\ \boxed{\begin{array}{l}\text{h is Planck’s constant }\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\text{ which relates energy and frequency}\\ \text{υ}\text{is the speed of particle}\\ \text{m is the mass of particle }\\ \text{λ}\text{is the wavelength}\\ \text{p is the momentum}\end{array}}\end{array}$

The above equation is called de Broglie relation.

Explanation of Solution

Given:

$10MeV$ Particle accelerator

The electron wavelength present in given accelerator is determined by using kinetic energy equation.

$\begin{array}{c}{\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{m}}_{\text{e}}{\text{v}}^{\text{2}}\\ \text{p}={\text{m}}_{\text{e}}\text{v}\\ {\text{E}}_{\text{k}}\text{=}\frac{{\text{p}}^{\text{2}}}{{\text{2m}}_{\text{e}}}\\ \text{p}=\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}\end{array}$

$\text{p}=\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}\to 1$

Substituting equation 1 in de Broglie equation

$\begin{array}{c}\text{p}=\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}\to 1\\ \text{λ=}\frac{\text{h}}{\text{p}}\Rightarrow \text{de}\text{Broglie}\text{equation}\\ =\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}}\to 2\end{array}$

For accelerated electron energy is equal to $e\Delta \varphi$. Then equation 2 becomes,

$\begin{array}{c}=\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}}\to 2\\ =\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}\text{eΔ}\varphi }}\left[{\text{E}}_{\text{k}}=\text{eΔ}\varphi \right]\\ \text{λ}=\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}\text{eΔ}\varphi }}\end{array}$

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}\text{eΔ}\varphi }}\\ =\frac{\text{6}\text{.626}\times {\text{10}}^{\text{-34}}}{\sqrt{\text{2}\left(9.109\times {10}^{-31}\right)\left(1.602\times {10}^{-19}\right)\left(1\times {10}^3\right)}}\times \frac{\text{J}\text{s}}{\sqrt{\text{kg}\text{C}\text{eV}}}\left[\text{1J=}\text{1}\text{CV=1}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\right]\\ =3.9\times {10}^{-11}\text{m}\\ \text{=39}\times {10}^{-12}\text{m}\\ \text{=}39\text{nm}\left[\text{1pm}\text{=}{\text{10}}^{\text{-11}}\text{m}\right]\\ \end{array}$

$\begin{array}{c}\text{p}=\frac{\text{h}}{\text{λ}}\\ =\frac{\text{6}\text{.626}\times {\text{10}}^{\text{-34}}}{39\times {10}^{-12}}\times \frac{\text{J}\text{s}}{\text{m}}\left[\text{1J=}\text{1}\text{CV=1}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\right]\\ =1.7\times {10}^{-23}\text{kg}\text{m}{\text{s}}^{\text{-1}}\\ \end{array}$

(c)

Interpretation Introduction

Interpretation:

The de Broglie wavelength for electron accelerated through $100kV$ and also the momentum for the electrons has to be calculated.

Concept Introduction:

Work function: Work function of a metal is measure of strength of electrons is held in a metal atom.

$\text{hν}\text{=}\text{KE}\text{+}\varphi \boxed{\begin{array}{c}here,\varphi =\text{ work function}\\ \text{KE }=\text{ kinetic energy}\\ \text{h = Planck’s constant} \\ \text{ν}=\text{ frequency}\end{array}}$

Energy can be in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.

${\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\boxed{\begin{array}{l}here,\text{m - Mass in kilograms}\hfill \\ \text{v - Velocity in meters per second}\hfill \end{array}}$

Louis de Broglie in $1923$ rationalized that when light shows particle aspects, then particles of matter display properties of waves under definite circumstances.

$\begin{array}{c}\text{λ}\text{=}\frac{\text{h}}{\text{m}\text{υ}}\\ \text{p = m}\text{υ}\\ \text{Hence λ}\text{=}\frac{\text{h}}{\text{p}}\\ \boxed{\begin{array}{l}\text{h is Planck’s constant }\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\text{ which relates energy and frequency}\\ \text{υ}\text{is the speed of particle}\\ \text{m is the mass of particle }\\ \text{λ}\text{is the wavelength}\\ \text{p is the momentum}\end{array}}\end{array}$

The above equation is called de Broglie relation.

Explanation of Solution

Given:

$10MeV$ Particle accelerator

The electron wavelength present in given accelerator is determined by using kinetic energy equation.

$\begin{array}{c}{\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{m}}_{\text{e}}{\text{v}}^{\text{2}}\\ \text{p}={\text{m}}_{\text{e}}\text{v}\\ {\text{E}}_{\text{k}}\text{=}\frac{{\text{p}}^{\text{2}}}{{\text{2m}}_{\text{e}}}\\ \text{p}=\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}\end{array}$

$\text{p}=\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}\to 1$

Substituting equation 1 in de Broglie equation

$\begin{array}{c}\text{p}=\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}\to 1\\ \text{λ=}\frac{\text{h}}{\text{p}}\Rightarrow \text{de}\text{Broglie}\text{equation}\\ =\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}}\to 2\end{array}$

For accelerated electron energy is equal to $e\Delta \varphi$. Then equation 2 becomes,

$\begin{array}{c}=\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}{\text{E}}_{\text{k}}}}\to 2\\ =\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}\text{eΔ}\varphi }}\left[{\text{E}}_{\text{k}}=\text{eΔ}\varphi \right]\\ \text{λ}=\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}\text{eΔ}\varphi }}\end{array}$

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{e}}\text{eΔ}\varphi }}\\ =\frac{\text{6}\text{.626}\times {\text{10}}^{\text{-34}}}{\sqrt{\text{2}\left(9.109\times {10}^{-31}\right)\left(1.602\times {10}^{-19}\right)\left(100\times {10}^3\right)}}\times \frac{\text{J}\text{s}}{\sqrt{\text{kg}\text{C}\text{eV}}}\left[\text{1J=}\text{1}\text{CV=1}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\right]\\ =3.88\times {10}^{-12}\text{m}\\ \text{=}3.88\text{pm}\left[\text{1pm}\text{=}{\text{10}}^{\text{-12}}\text{m}\right]\end{array}$

$\begin{array}{c}\text{p}=\frac{\text{h}}{\text{λ}}\\ =\frac{\text{6}\text{.626}\times {\text{10}}^{\text{-34}}}{3.88\times {10}^{-12}}\times \frac{\text{J}\text{s}}{\text{m}}\left[\text{1J=}\text{1}\text{CV=1}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\right]\\ =1.71\times {10}^{-22}\text{kg}\text{m}{\text{s}}^{\text{-1}}\\ \end{array}$