Explanation Given data: The atom whose ground state energy is to be determined is hydrogen. Formula used: According to Bohr's postulate, the angular momentum L for stable electron’s orbit is quantized. Mathematically, the expression for the quantized angular momentum is: L = m v r = n h 2 π Here, n = 1 , 2 , 3 , .... represents the integer value, m is the mass of the electron, v is the velocity of the electron, r is the radius of the orbit, n is the principal quantum number, and h is Planck's constant. The expression for the centripetal force required for the circular motion of the electron around the nucleus is: F C = m v 2 r Here, m is the mass of the electron, v is the velocity of electron and r is the radius of the orbit. The expression for the electrostatic force between electron and nucleus is: F E = k e 2 r 2 Here, r is the radius of the orbit, k is Coulomb's constant ( k = 9 × 10 9 N ⋅ m 2 ⋅ C − 2 ) and e is the electronic charge of the electron ( e = 1.6 × 10 9 C ). The expression for the kinetic energy of an atom is: K = 1 2 m v 2 Here, m is the mass and v is the speed of the atom. The expression for the potential energy of an atom is: U = − k e 2 r Here, k is Coulomb’s constant, e is the electronic charge of the atom and r is the radius of the orbit. Explanation: Recall the expression for the angular momentum. L = m v r = n h 2 π Solving for v . v = n h 2 π m r ……(1) In a hydrogen atom, the required centripetal force for the circular motion of the electron is provided by the electrostatic force between the electron and the nucleus. So, F C = F E Substitute m v 2 r for F C and k e 2 r 2 for F E . m v 2 r = k e 2 r 2 Rearrange the expression for v : v = ( k e 2 m r ) 1 2 …… (2) From equation (1) and (2). r = h 2 4 π 2 k e 2 m n 2 Substitute h 2 4 π 2 k e 2 m n 2 for r in equation (1). v = 2 π k e 2 h 1 n The total energy of a hydrogen atom is the sum of electric potential energy U plus electron’s kinetic energy K . So, E = U + K Substitute − k e 2 r for U and 1 2 m v 2 for K

College Physics: Explore And Apply, Volume 2 (2nd Edition)

2nd Edition

ISBN: 9780134862910

Author: Eugenia Etkina, Gorazd Planinsic, Alan Van Heuvelen, Gorzad Planinsic

Publisher: PEARSON

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Chapter 28, Problem 6P

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The way to calculate that the energy of a hydrogen atom in its ground state is −13.6 eV.

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3.37(a) Five free electrons exist in a three-dimensional infinite potential well with all three widths equal to \( a = 12 \, \text{Å} \). Determine the Fermi energy level at \( T = 0 \, \text{K} \). (b) Repeat part (a) for 13 electrons. Book: Semiconductor Physics and Devices 4th ed, NeamanChapter-3Please expert answer only. don't give gpt-generated answers, & please clear the concept of quantum states for determining nx, ny, nz to determine E, as I don't have much idea about that topic.

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Chapter 28 Solutions

College Physics: Explore And Apply, Volume 2 (2nd Edition)

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The way to calculate that the energy of a hydrogen atom in its ground state is − 13.6 eV .

Solution Summary: The author explains Bohr's postulate that the energy of a hydrogen atom in its ground state is -13.6eV.

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The way to calculate that the energy of a hydrogen atom in its ground state is −13.6 eV.

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Chapter 28 Solutions