(a) Explanation Write the equation to find the speed of electrons in the n t h orbit of hydrogen atom. v n = 1 ε 0 e 2 2 n h (1) Here, v n is the radius of the n t h orbit of hydrogen. ε 0 is the permittivity of free space. e is the charge of electron. n is the principle quantum number. h is the planck’s constant. Substitute 1 for n , 6.626 × 10 − 34 Js for h , 1.6 × 10 − 19 C for e , and 8.987 × 10 9 N m 2 / C 2 for ε 0 in equation (1). v 1 = ( 1.60 × 10 − 19 C ) 2 8.987 × 10 9 N m 2 / C 2 ×2×6 .626×10 -34 Js = 2.19 × 10 6 m / s Write the value for the orbital radii r n r n = 5.291 × 10 − 11 m Write the expression for the kinetic energy K n of the electron. K n = 1 2 m v n 2 (2) Here, K is the kinetic energy of the electron. m is the mass of the electron Substitute 9.1 × 10 − 31 kg for m and 2.19 × 10 6 m / s in equation (2). K 1 = 1 2 ( 9.1 × 10 − 31 kg ) ( 2.19 × 10 6 m / s ) 2 = 2.185 × 10 − 18 J 1 e V 1.6 × 10 − 19 J = 13 (b) To determine For any value of n , both potential and kinetic are U n = − 2 K n and K n = − E n respectively.

Mastering Physics with Pearson eText -- Standalone Access Card -- for University Physics with Modern Physics (14th Edition)

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ISBN: 9780133978216

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Question

Chapter 39, Problem 39.24E

(a)

To determine

The total energy, the kinetic energy and the potential energy for energy level n=1 of hydrogen atom.

(b)

To determine

For any value of n, both potential and kinetic are Un=−2Kn and Kn=−En respectively.

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Chapter 39, Problem 39.23E

Chapter 39, Problem 39.25E

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Chapter 39 Solutions

Mastering Physics with Pearson eText -- Standalone Access Card -- for University Physics with Modern Physics (14th Edition)

Ch. 39.2 - Prob. 39.2TYU Ch. 39.3 - Prob. 39.3TYU Ch. 39.4 - Prob. 39.4TYU Ch. 39.5 - Prob. 39.5TYU Ch. 39.6 - Prob. 39.6TYU Ch. 39 - Prob. 39.1DQ Ch. 39 - Prob. 39.2DQ Ch. 39 - Prob. 39.3DQ Ch. 39 - When an electron beam goes through a very small...Ch. 39 - Prob. 39.5DQ

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The total energy, the kinetic energy and the potential energy for energy level n = 1 of hydrogen atom.

Solution Summary: The author explains the equation to find the speed of electrons in the n_th orbit of hydrogen atom.

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The total energy, the kinetic energy and the potential energy for energy level n=1 of hydrogen atom.

For any value of n, both potential and kinetic are Un=−2Kn and Kn=−En respectively.

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Chapter 39 Solutions