The Bohr Model The hydrogen atom consists of one electron and one proton. In the

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.

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