Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Using the classical equations for momentum and kinetic energy, show that an electron’s de Broglie wavelength in nanometres can be written as $λ = 1.226 / \sqrt{K}$ , in which K is the electron’s kinetic energy in electron-volts.

The required expression for the de Broglie wavelength is, $\frac{1.226 \times 10^{- 9} {m.eV}^{\frac{1}{2}}}{\sqrt{K}}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the given data from the question

The classical equation of momentum is, $p = m v$

The kinetic energy is, $K = \frac{1}{2} m v^{2}$

Step 2: Determine the formulas to prove the de Broglie wavelength λ=1.226/K

The expression to calculate the kinetic energy of the electron is given as follows.

$K = \frac{1}{2} m v^{2}$ …(i)

Here, m is the mass of the electron and v is the velocity of electron.

The expression to calculate the de Broglie wavelength is given as follows.

$λ = \frac{h}{p}$ …(ii)

Here, h is the plank’s constant and $p$ is the momentum.

Step 3: Prove the Broglie wavelength λ=1.226/K

Calculate the kinetic energy of the electron,

Substitute $ρ / m$ for v into equation (i).

$\begin{array}{rcl} K & = & \frac{1}{2} m {(\frac{p}{m})}^{2} \\ K & = & \frac{1}{2} \frac{p^{2}}{m} \\ p^{2} & = & 2 K m \\ p & = & \sqrt{2 K m} \end{array}$

Calculate the de Broglie wavelength.

Substitute $\sqrt{2 K m}$ for $p$ into equation (ii).

$λ = \frac{h}{\sqrt{2 K m}}$

Substitute $6.62 \times 10^{- 34} Js$ for h and $9.11 \times 10^{- 31} kg$ for m into above equation.

$\begin{array}{rcl} λ & = & \frac{6.62 \times 10^{- 34} Js}{\sqrt{2 \times (9.11 \times 10^{- 31} kg) \times (1.602 \times 10^{- 19} J / eV) \times K}} \\ = & \frac{6.62 \times 10^{- 34} Js}{\sqrt{29.1884 \times 10^{- 50} kgJ / eV}} \frac{1}{\sqrt{K}} \\ = & \frac{1.226 \times 10^{- 9} {meV}^{\frac{1}{2}}}{\sqrt{K}} \end{array}$

Hence the required expression for the de Broglie wavelength is,

\begin{array}{rcl} \frac{1.226 \times 10^{- 9} {meV}^{\frac{1}{2}}}{\sqrt{K}} \end{array}

	Electron Energy	Barrier Height	Barrier Thickness
(a)	E	5E	L
(b)	E	17E	L/2
(c)	E	2E	2L

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Fundamentals Of Physics

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Short Answer

Using the classical equations for momentum and kinetic energy, show that an electron’s de Broglie wavelength in nanometres can be written as $λ = 1.226 / \sqrt{K}$ , in which K is the electron’s kinetic energy in electron-volts.

Step by Step Solution

Step 1: Write the given data from the question

Step 2: Determine the formulas to prove the de Broglie wavelength λ=1.226/K

Step 3: Prove the Broglie wavelength λ=1.226/K

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Fundamentals Of Physics

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Short Answer

Using the classical equations for momentum and kinetic energy, show that an electron’s de Broglie wavelength in nanometres can be written as λ=1.226/K, in which K is the electron’s kinetic energy in electron-volts.

Step by Step Solution

Step 1: Write the given data from the question

Step 2: Determine the formulas to prove the de Broglie wavelength λ=1.226/K

Step 3: Prove the Broglie wavelength λ=1.226/K

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Using the classical equations for momentum and kinetic energy, show that an electron’s de Broglie wavelength in nanometres can be written as $λ = 1.226 / \sqrt{K}$ , in which K is the electron’s kinetic energy in electron-volts.