Explanation 1) Concept: We find the total energy of the electron and the wave number in region 1. Finding the value of b and substituting in transmission coeffient formula, we can find the transmission coefficient of the barrier. Taking the product of transmission coefficient and the initial current of region 1, we can find the transmitted current in region 2. 2) Formula: i) E = 1 2 m v 2 ii) k = 2 π λ iii) T = e - 2 b L iv) b = 8 π 2 m U b - E h 2 v) T = I t I 0 3) Given: i) Speed of electrons v = 1.200 × 10 3 m s ii) Current I 0 = 9.000 m A iii) Height of potential barrier U b = - 4.719 μ V iv) Thickness L = 200.0 n m 4) Calculations: The energy of the electron in the region 1 is E = K + P . E E = 1 2 m v 2 + 0 E = 1 2 ( 9.11 × 10 - 31 k g ) 1.2 × 10 3 m s 2 E = 6.56 × 10 - 25 J E = 4.0995 μ e V The wave number in region 1 is k = 2 π λ The de-Broglie wavelength is λ = h p k = 2 π p h The linear momentum p = m v k = 2 π m v h k = 2 π ( 9.11 × 10 - 31 k g ) ( 1200 m s ) 6

Fundamentals of Physics, Volume 1, Chapter 1-20

10th Edition

ISBN: 9781118233764

Author: David Halliday

Publisher: WILEY

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Chapter 38, Problem 78P

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Transmitted current.

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Chapter 38, Problem 77P

Chapter 38, Problem 79P

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The current of a beam of electrons, each with a speed of 1.200*10^3 m/s, is 9.000 mA. At one point along its path, the beam encounters a potential barrier of height -4.719 mV and thickness 200.0 nm.What is the transmitted current?

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Chapter 38 Solutions

Fundamentals of Physics, Volume 1, Chapter 1-20

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Transmitted current.

Solution Summary: The author calculates the energy of the electron in the region 1 and the wave number in region 1.

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To find: Transmitted current.

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Chapter 38 Solutions

To find:

Transmitted current.