The Bohr Model The hydrogen atom consists of one electron and one proton. In the
ID: 3894240 • Letter: T
Question
The Bohr Model The hydrogen atom consists of one electron and one proton. In the Bohr model of the hydrogen atom, the electron orbits the proton in a circular orbit.Part A What is the smallest amount of work that must be done on the electron to move it from its circular orbit, with a radius of 0.529 The Bohr Model The hydrogen atom consists of one electron and one proton. In the Bohr model of the hydrogen atom, the electron orbits the proton in a circular orbit.
The Bohr Model The hydrogen atom consists of one electron and one proton. In the Bohr model of the hydrogen atom, the electron orbits the proton in a circular orbit.
The Bohr Model The hydrogen atom consists of one electron and one proton. In the Bohr model of the hydrogen atom, the electron orbits the proton in a circular orbit.
Part A What is the smallest amount of work that must be done on the electron to move it from its circular orbit, with a radius of 0.529
Explanation / Answer
There are two ways to go about this. One is based on electrostatic charge and orbital mechanics, and the other on the Bohr formula.
Electrostatic-orbital first, since the radius is the only data given.
Orbital mechanics says the total energy PE+KE of a body in circular orbit equals 1/2 the potential energy PE.
Electrostatics says PE = kq1q2/r.
We know q1 = -q2 (electron) = 1.6021774E-19C and k = 8.98755E9 N-m^2/C^2, so PE = -8.98755E9*1.6021774E-19^2/0.529E-10 = -4.3612085E-18 J.
Dividing by 2, E = -2.1806043E-18 J or 13.610255 eV.
Bohr formula next.
This is easier because it basically encapsulates the electrostatic-orbital method (see refs.).
It uses the ground-state energy Eg = -2.17987197E-18 J-N^2, in the formula E = Eg/N^2. Assuming the orbit given is that for the ground state (N=1),
the answer is just E = -2.17987197E-18 J, or -13.605692 eV.
Both methods agree very well, especially considering that the radius is given to only 3 sig. figs. This agreement also verifies that the given radius is that of the ground state.
Note that since PE = 0 at infinite distance, and in the case of opposite-polarity (attracting) charges is negative at closer distances, all the energies given are negative (although the minus sign is often omitted). Thus the work to ionize the atom is positive.