Explanation The value of Planck’s constant is 6.626 × 10 − 34 Kg ⋅ m 2 ⋅ s − 1 . The value of Coulomb’s constant is 8.988 × 10 9 N ⋅ m 2 ⋅ C − 2 . The charge on an electron is 1.6 × 10 − 19 C . The mass of an electron is 9.11 × 10 − 31 Kg . The value of π is 3.14 . The relation between N and Kg ⋅ m ⋅ s − 2 is, 1 N = 1 Kg ⋅ m ⋅ s − 2 Use this relation to convert given value of Coulomb’s constant. 8.988 × 10 9 N ⋅ m 2 ⋅ C − 2 = 8.988 × 10 9 Kg ⋅ m 3 ⋅ s − 2 ⋅ C − 2 Formula to calculate the de Broglie wavelength of an electron ( λ ) in the ground state of the hydrogen atom is, λ = h m v Here, h is the Planck’s constant. m is mass of electron. v is velocity of an electron of hydrogen atom placed at the ground state. Formula to calculate the velocity of an electron ( v n ) in nth orbit of hydrogen atom is, v n = 2 πke 2 n h Here, k is the Coulomb’s constant. e is charge on electron. n is number of orbits. h is the Planck’s constant. π is constant. The value of n for ground state of hydrogen atom is 1 . Substitute 1 for n to find the velocity of an electron ( v ) placed at ground state of hydrogen atom. v = 2 πke 2 h Substitute 6.626 × 10 − 34 Kg ⋅ m 2 ⋅ s − 1 for h , 8.988 × 10 9 Kg ⋅ m 3 ⋅ s − 2 ⋅ C − 2 for k , 1.6 × 10 − 19 C for e and 3

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Chapter 31, Problem 35PCE

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The de Broglie wavelength of an electron in the ground state of the hydrogen atom.

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Chapter 31, Problem 34PCE

Chapter 31, Problem 36PCE

Chapter 31 Solutions

Physics (5th Edition)

Ch. 31.1 - Prob. 1EYU Ch. 31.2 - Prob. 2EYU Ch. 31.3 - Prob. 3EYU Ch. 31.4 - Prob. 4EYU Ch. 31.5 - Prob. 5EYU Ch. 31.6 - Prob. 6EYU Ch. 31.7 - Prob. 7EYU Ch. 31 - Prob. 1CQ Ch. 31 - Prob. 2CQ Ch. 31 - Prob. 3CQ

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