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(a) Write an expression relating the kinetic energy $K E$ of the electron and the potential energy $P E$ in the Bohr model of the hydrogen atom. (b) Suppose a hydrogen atom absorbs a photon of energy $E,$ resulting in the transfer of the electron to a higher-energy level. Express the resulting change in the potential energy of the system in terms of $E$ . (c) What is the change in the electron's kinetic energy during this process?

a) $K E_{n}=-\frac{1}{2} P E$

b) $\Delta P E=2 E$

c) $\Delta K E=-E$

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Rutgers, The State University of New Jersey

Numerade Educator

University of Sheffield

McMaster University

in question. A. We have to find an equation that relates the kinetic energy of an electric and the potential energy of the system when an electric is orbiting an atomic nucleus. So we know that the kinetic energy K e often electric with is just the mass of the electron times, the speed square over to and the potential energy of the system is columns constant times The charge of the election, which is minus e times the charge of the broken, which is just plus e divided by the radius of the orbit eso he way have to do here is to somehow find equations for the Speed V and for the radius are such that we can relate the kinetic energy of the potential energy. Luckily from the boar model, we have that the radius of the of the the orbit of the election is going by and squared times the Bohr radius, which is H bar squared, divided by K him he squared where m is the mass of the election and also we have that the angular momentum l is equal to end h bar and also you know that the angular momentum can be expressed as M v R. Okay, so what I'm gonna do year is to isolate are in this for Arabs are addressing a TV in this formula, and I'm gonna substitute it back into equation, um, into the equation for the Connecticut knows you so have that an H bar is equal to I am times Wien Times are are is what, uh, what we found here in the first equation. So I'm gonna Just so city are, which is in squared h bar squared, divided by K and E square. So the EMS gets allowed the end an H bar. One of them cancels out. Yes, So we're left with V equals Okay, E square divided by in a church. Mom and I can substitute it back in the Kanada candidate equation. So the kinetic Canada's you is I m times v square. So the square it is d squared into the fourth and squared h Worse, where the right invite to and the potential energy potential energy P e is equal to minus que e squared, divided by R And as you saw our is people to end swear h bar squared, divided by K and E square So this is minus K e squared, divided by and square each square. Uh, times okay in he square. So this here is minus case where I am e to the fourth divided by and square H Bar Square. So notice that this is exactly minus one minus two times de kinetic energy. Okay, so this is the relation between the potential energy p e and the kinetic energy. K e in question be we have a fault that's absorbed by the by the Adam, and we want to calculate what's the difference? That's a P e. In the potential energy of the atom, due to the absorption of the Fulton as a function of the energy of the photon e, He is the energy of absorbed Fulton. And we want to find Dr P so noticed that the change in energy off the Adam the total change in energy is just the energy of the Fulton by conservation of energy. Okay. But we also have that the total change in energy is a change in the potential energy. That's a P plus the change in the kinetic energy belt, a k E. And we have that not a K e way No, from compression A. You know that the potential energy is minus two times the kinetic energy. So the change in the potential energy and I'm going to write it here in blue The change in the potential energy is ma I'm sorry is minus two times the change in the kinetic energy. So the change of magnetic energy is miners. 1/2 of the change in the potential energy. So I can apply this in and formula for those of you. So I have that the p e miners 1/2 off Delta Bi So I have about a e is people too. 1/2 of Delta p E. But after he is able to the energy off the being coming for Eve, so have the to be the difference in the potential. Energy is two times the the energy of the incoming photo. Okay, and questions. See, you have to calculate the change in the kinetic energy and as I showed here in blue, the difference in the kinetic energy is mine is 1/2 of the difference in the potential energy. So since the difference in the potential energy is to be the difference in the kinetic energy you is minus. He would be that minors 1/2 of that OPD. So that concludes the exercise.

Universidade de Sao Paulo