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

2nd Edition

ISBN: 9780134601823

Author: ETKINA, Eugenia, Planinšič, G. (gorazd), Van Heuvelen, Alan

Publisher: Pearson,

See similar textbooks

Concept explainers

Quantum mechanics and hydrogen atom

Consider an electron of mass m moves with the velocity v in a hydrogen atom. If an electron is at a distance r from the proton, then the potential energy function of the electron can be written as follows:

Isotopes of Hydrogen Atoms

To understand isotopes, it's easiest to learn the simplest system. Hydrogen, the most basic nucleus, has received a great deal of attention. Several of the results seen in more complex nuclei can be seen in hydrogen isotopes. An isotope is a nucleus of t…

Mass of Hydrogen Atom

Hydrogen is one of the most fundamental elements on Earth which is colorless, odorless, and a flammable chemical substance. The representation of hydrogen in the periodic table is H. It is mostly found as a diatomic molecule as water H2O on earth. It is …

Question

Chapter 28, Problem 6P

To determine

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 28, Problem 5P

Chapter 28, Problem 7P

Students have asked these similar questions

3) If the slider block C is moving at 3m/s, determine the angular velocity of BC and the crank AB at the instant shown. (Use equation Vs Vc wxf, then use equation V, VA + Va/Athen write it in terms of w and the appropriate r equate the two and solve) f-3m

Pls help ASAP

Chapter 28 Solutions

College Physics

Ch. 28 - Prob. 1RQ Ch. 28 - Prob. 2RQ Ch. 28 - Prob. 3RQ Ch. 28 - Prob. 4RQ Ch. 28 - Prob. 5RQ Ch. 28 - Prob. 6RQ Ch. 28 - Prob. 7RQ Ch. 28 - Prob. 8RQ Ch. 28 - Prob. 1MCQ Ch. 28 - Prob. 2MCQ

Ch. 28 - Prob. 3MCQ Ch. 28 - Prob. 4MCQ Ch. 28 - Prob. 5MCQ Ch. 28 - Prob. 6MCQ Ch. 28 - Prob. 7MCQ Ch. 28 - Prob. 8MCQ Ch. 28 - Prob. 9MCQ Ch. 28 - Prob. 10MCQ Ch. 28 - Prob. 11MCQ Ch. 28 - Prob. 12MCQ Ch. 28 - Prob. 13CQ Ch. 28 - Prob. 14CQ Ch. 28 - Prob. 15CQ Ch. 28 - Prob. 16CQ Ch. 28 - Prob. 17CQ Ch. 28 - Prob. 18CQ Ch. 28 - Prob. 19CQ Ch. 28 - Prob. 20CQ Ch. 28 - Prob. 21CQ Ch. 28 - Prob. 22CQ Ch. 28 - Prob. 23CQ Ch. 28 - Prob. 24CQ Ch. 28 - Prob. 25CQ Ch. 28 - Prob. 26CQ Ch. 28 - Prob. 27CQ Ch. 28 - Prob. 28CQ Ch. 28 - Prob. 29CQ Ch. 28 - Prob. 30CQ Ch. 28 - Prob. 31CQ Ch. 28 - Prob. 32CQ Ch. 28 - Prob. 33CQ Ch. 28 - Prob. 34CQ Ch. 28 - Prob. 1P Ch. 28 - Prob. 2P Ch. 28 - Prob. 3P Ch. 28 - Prob. 4P Ch. 28 - Prob. 5P Ch. 28 - Prob. 6P Ch. 28 - Prob. 7P Ch. 28 - Prob. 8P Ch. 28 - Prob. 9P Ch. 28 - Prob. 10P Ch. 28 - Prob. 11P Ch. 28 - Prob. 12P Ch. 28 - Prob. 13P Ch. 28 - Prob. 14P Ch. 28 - Prob. 15P Ch. 28 - Prob. 16P Ch. 28 - Prob. 17P Ch. 28 - Prob. 18P Ch. 28 - Prob. 19P Ch. 28 - Prob. 20P Ch. 28 - Prob. 21P Ch. 28 - Prob. 22P Ch. 28 - 28.4 Lasers (a) A laser pulse emits 2.0 J of...Ch. 28 - Prob. 24P Ch. 28 - Prob. 25P Ch. 28 - Prob. 26P Ch. 28 - Prob. 27P Ch. 28 - Prob. 28P Ch. 28 - Prob. 29P Ch. 28 - Prob. 30P Ch. 28 - Prob. 31P Ch. 28 - Prob. 32P Ch. 28 - Prob. 33P Ch. 28 - Prob. 34P Ch. 28 - Prob. 35P Ch. 28 - Prob. 36P Ch. 28 - Prob. 37P Ch. 28 - Prob. 38P Ch. 28 - Prob. 39P Ch. 28 - Prob. 40P Ch. 28 - Prob. 41P Ch. 28 - Prob. 42P Ch. 28 - Prob. 43P Ch. 28 - Prob. 44P Ch. 28 - Prob. 45P Ch. 28 - Prob. 46P Ch. 28 - Prob. 47P Ch. 28 - Prob. 48P Ch. 28 - Prob. 49P Ch. 28 - Prob. 50P Ch. 28 - Prob. 51P Ch. 28 - Prob. 52P Ch. 28 - Prob. 53P Ch. 28 - Prob. 54P Ch. 28 - Prob. 55P Ch. 28 - Prob. 56P Ch. 28 - Prob. 57P Ch. 28 - Prob. 58P Ch. 28 - Prob. 59GP Ch. 28 - Prob. 60GP Ch. 28 - Prob. 61GP Ch. 28 - Prob. 62GP Ch. 28 - Prob. 63GP Ch. 28 - Prob. 64GP Ch. 28 - Prob. 65GP Ch. 28 - Prob. 66GP Ch. 28 - Prob. 67GP Ch. 28 - Prob. 68RPP Ch. 28 - Prob. 69RPP Ch. 28 - Prob. 70RPP Ch. 28 - Prob. 71RPP Ch. 28 - Prob. 72RPP Ch. 28 - Prob. 73RPP Ch. 28 - Prob. 74RPP Ch. 28 - Prob. 75RPP Ch. 28 - Prob. 76RPP Ch. 28 - Prob. 77RPP Ch. 28 - Prob. 78RPP

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

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.

Concept explainers

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

Want to see the full answer?

Chapter 28 Solutions