Explanation Given data: Allowed energy levels are 0.0 eV, 4.0 eV and 6.0 eV. Speed of the electron is 1.6 × 10 6 m s . Formula used: Write the expression for kinetic energy. K = 1 2 m v 2 (1) Here, K is the kinetic energy, m is the mass of the electron with value 9.11 × 10 − 31 kg , and v is the velocity. Explanation: Substitute 9.11 × 10 − 31 kg for m and 1.6 × 10 6 m s for v in equation (1), K = 1 2 × ( 9.11 × 10 − 31 kg ) × ( 1.6 × 10 6 m s ) 2 = ( 2.332 × 10 − 18 2 ) J =1 .166 × 10 − 18 J Consider the conversion factor for J to eV. 1 J = ( 1 1.6 × 10 − 19 ) eV Convert the Kinetic energy in eV, K = ( 1 .166 × 10 − 18 × 1 1.6 × 10 − 19 ) eV = 7 .2875 eV ≅ 7 .29 eV For minimum speed, The electron travelling to third energy level that is 6.0 eV will be left with 7.29 eV − 6.0 eV = 1.29 eV energy. Left over kinetic energy in J, K = ( 1.29 eV × ( 1.6 × 10 − 19 1 eV ) ) J = 2 .064 × 10 − 19 J Substitute 9.11 × 10 − 31 kg for m and 2 .064 × 10 − 19 J for K in equation (1), 2 .064 × 10 − 19 J = 1 2 × ( 9.11 × 10 − 31 kg ) × v 2 v 2 = 2

Student Workbook For College Physics Format: Paper

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ISBN: 9780134724805

Author: Knight, Randall D., (professor Emeritus)^jones, Br

Publisher: Prentice Hall

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The Bohr model

Students often get confused between Rutherford’s model and Bohr model of an atom. They interpret the postulates of both these models to be the same, which is incorrect. Rutherford’s model differs from Bohr’s model in the following ways:

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Chapter 29, Problem 17P

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The minimum and maximum speeds of the electron after collision.

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Chapter 29, Problem 16P

Chapter 29, Problem 18P

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