EBK PHYSICS FOR SCIENTISTS & ENGINEERS
5th Edition
ISBN: 9780134296074
Author: GIANCOLI
Publisher: VST
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Chapter 37, Problem 77GP
To determine
The ratio of the gravitational force to the electric force for the electron in a hydrogen atom and to check whether the gravitational force can be ignored.
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Say you have a point charge of one Coulomb and a hydrogen atom some distance away. Within what distance would the difference of forces on the proton and electron from the Coulomb charge exceed the attraction between the proton and electron that holds the hydrogen atom together. Treat the hydrogen atom as a proton and electron 0.53 × 10−10 m (a.k.a., one Bohr radius) apart.
Determine the distance between the electron and proton in an atom if the potential energy U of the electron is 13.8 ev (electronvolt, 1 eV = 1.6 × 10-19 J). Give your answer in
Angstrom (1 A = 10-10 m).
Answer:
Choose... +
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An electron revolves around the nucleus of an atom
in a circular orbit of radius
4.0Å with a speed of 6.0 x 10^6 ms-1. Calculate the
linear kinetic energy.
Chapter 37 Solutions
EBK PHYSICS FOR SCIENTISTS & ENGINEERS
Ch. 37 - Prob. 1QCh. 37 - Prob. 2QCh. 37 - Prob. 3QCh. 37 - Prob. 4QCh. 37 - Prob. 5QCh. 37 - Prob. 6QCh. 37 - Prob. 7QCh. 37 - Prob. 8QCh. 37 - Prob. 9QCh. 37 - Prob. 10Q
Ch. 37 - Prob. 11QCh. 37 - Prob. 12QCh. 37 - Prob. 13QCh. 37 - Prob. 14QCh. 37 - Prob. 15QCh. 37 - Prob. 16QCh. 37 - Prob. 17QCh. 37 - Prob. 18QCh. 37 - Prob. 19QCh. 37 - Prob. 20QCh. 37 - Prob. 1PCh. 37 - Prob. 3PCh. 37 - Prob. 7PCh. 37 - Prob. 9PCh. 37 - Prob. 13PCh. 37 - Prob. 25PCh. 37 - Prob. 26PCh. 37 - Prob. 27PCh. 37 - Prob. 32PCh. 37 - Prob. 34PCh. 37 - Prob. 37PCh. 37 - Prob. 45PCh. 37 - Prob. 52PCh. 37 - Prob. 76GPCh. 37 - Prob. 77GP
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- The force on an electron is “negative the gradient of the potential energy function.” Use this knowledge and Equation 8.1 to show that the force on the electron in a hydrogen atom is given by Coulomb’s force law. Ur=ke2r(8.1)arrow_forwardIn the simple Bohr model of the ground state of the hydrogen atom, the electron travels in a circular orbit around a fixed proton. The radius of the orbit is 5.281011m , and the speed of the electron is 2.18106m/s . The mass of an electron is 9.111031kg . What is the force on the electron?arrow_forwardDetermine the distance between the electron and proton in an atom if the potential energy U of the electron is 13.8 ev (electronvolt, 1 eV -19 1.6 x 10 J). Give your answer in Angstrom (1 A = 10"1º m). Answer: Choose... + Next pagearrow_forward
- 8. In a simple model of the hydrogen atom, an electron of mass m and charge e is considered to move in a nearly circular orbit about a proton. (i) Write down the expression for the force on the electron, and show that the kinetic energy of the electron is where r is the radius of the e2 περι orbit and ε, is the permittivity of free space. (ii) Find the total energy of the electron. nh (iii) Given that the angular momentum of the electron equal to where n is an integer and h is Plank's constant, show that the total energy of the electron is: En =- - me 8ɛh²/ n²arrow_forwardThe electron of a hydrogen atom is at a distance 5.3x10 m from the nucleus. If the charge of the nucleus is 1.6x10¹9 C, then the electrostatic potential energy (in electron volt (eV)) of the electron is: Select one: a. -1.5 b. 27.2 c. 1.5 d. -27.2 e. zeroarrow_forwardCoulomb's Law yields an expression for the energy of interaction for a pair of point charges. V = 2.31x10^-19 Q1Q2 r V is the energy (in J) required to bring the two charges from infinite distance separation to distance r (in nm).Q1 and Q2 are the charges in terms of electrons.(i.e. the constant in the above expression is 2.31×10-19 J nm electrons-2) For a group of "point" charges (e.g. ions) the total energy of interaction is the sum of the interaction energies for the individual pairs.Calculate the energy of interaction for the square arrangement of ions shown in the diagram below. d = 0.545 nmarrow_forward
- The nucleus of a certain atom is 7.30 fm in diameter and has 76.0 protons. What is the acceleration of another proton 4.00 fm from the surface of the nucleus? - 2.35×1027 m/s2, away from the nucleus- 8.20×1028 m/s2, towards the nucleus- 6.55×1029 m/s2, towards the nucleus- 1.79×1029 m/s2, away from the nucleusarrow_forward. In the Bohr model of the hydrogen atom, an electron in the lowest energy state follows a circular path. [ Charge (q) = 1.6 x 10-19 C Coulomb's constant (k) = 8.89 x 10° N.m²/C?. Mass of electron (me) = 9.109 x 10-31 kg Plank's constant (h) = 6.63 x 10 34 J.s Bohr radius (ao) = 0.0529 nm] a) What distance does an electron in the ground state travel? b) Determine the velocity of the electron around its orbit. c) What is the effective current associated with this orbiting electron? %3D %3Darrow_forward1.00g of hydrogen contains 6.02x10^23 atoms, each with one electron and one proton. Suppose that 1.00g of hydrogen is separated into protons and electrons, that the protons are placed at Earth's north pole, and that electrons are placed at Earth's south pole. Find the magnitude of the resulting compressional force on Earth. (The radius of Earth is approximately 6.38x10^6 m).arrow_forward
- The classic Millikan oil drop experiment setup is shown below. In this experiment oil drops are suspended in a vertical electric field against the gravitational force to measure their charge. If the mass of a negatively charged drop suspended in an electric field of 1.76 ✕ 10−4 N/C strength is 8.620 ✕ 10−21 g, find the number of excess electrons in the drop.arrow_forwardAn electron is 0.5 Å away from a carbon nucleus which contains 6 protons. Find the mutual force of attraction between the nucleus and the electron. 1Å (Angstrom)= 10^-10m, e=1.6x10^-19C.arrow_forward1.00g of hydrogen contains 6.02x10^23 atoms, each with one electron and one proton. Suppose that 1.00g of hydrogen is separated into protons and electrons, that the protons are placed at Earth's north pole, and that electrons are placed at Earth's south pole. Find the magnitude of the resulting compressional force on Earth. (The radius of Earth is approximately 6.38x10^6 m). * 5.12x10^5 N 4.23x10^5 N 6.43x10^6 N 6.87x10^6 N None of the Abovearrow_forward
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