Explanation Given Info: The electron in a hydrogen atom considered as a circular orbit. The radius of the circular orbit is 0.0529 nm . The kinetic energy is 13.6 eV . The charge of the electron is 1.60 × 10 − 19 C . The value of 1eV is 1.6 × 10 − 19 J The mass of the electron is 9.109 × 10 − 31 kg . Formula for the rate at which energy is emitted from an acceleration charge q and the acceleration a is: d E d t = q 2 a 2 6 π ε 0 c 3 Formula to calculate the acceleration of the circular orbit, a = v 2 R (I) v is the velocity of the circular orbit, R is the radius of the circular orbit. Multiply numerator and denominator with 1 2 m in equation (I), a = 1 2 m v 2 1 2 m R (II) Kinetic energy of the electron is ( 13.6 eV ) . K E = 13.6 eV The value of 1eV is 1.6 × 10 − 19 J / eV for ( 13.6 eV ) is, K E = ( 13.6 eV ) ( 1.6 × 10 − 19 J / eV ) = 2.176 × 10 − 18 J Substitute K E for 1 2 m v 2 in equation (II), a = K E 1 2 m R = 2 K E m R Substitute 2.176 × 10 − 18 J for K E , 9.109 × 10 − 31 kg for m and 0.0529 × 10 − 9 m to find a , a = 2 ( 2.176 × 10 − 18 J ) ( 9.109 × 10 − 31 kg ) ( 0.0529 × 10 − 9 m ) = 9.031 × 10 22 m / s 2 Thus, the acceleration of the circular orbit is 9.031 × 10 22 m / s 2 . Formula to calculate the rate of energy emission for the hydrogen atom is, d E d t = q 2 a 2 6 π ε 0 c 3 q is the acceleration charge, a is the acceleration of the hydrogen atom, ε 0 is the electric constant, c is the speed of the light. Substitute 1.60 × 10 − 19 C for q , 9.031 × 10 22 m / s 2 for a , 8.854 × 10 − 12 C 2 / N ⋅ m 2 for ε 0 and 2.998 × 10 8 m / s for c to find d E d t , d E d t = ( 1

University Physics with Modern Physics (14th Edition)

14th Edition

ISBN: 9780321973610

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Chapter 32, Problem 32.52CP

To determine

The energy required for the hydrogen atom in radiate per second.

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Chapter 32, Problem 32.51CP

Chapter 32, Problem 32.53CP

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University Physics with Modern Physics (14th Edition)

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