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I came across the following part in my text book. Bohr's atomic theory explained the stability of an atom by the idea of energy shells. He said that while orbiting the nucleus through a particular orbit or energy shell the energy of electrons neither increases nor decreases. The electron does not carry energies while orbiting. The shells are charged and the energy of electrons orbiting through the shell equals to the energy of the shell and as the shells are stable therefore the energy of electrons are also stable . Now, the energy increases from shell to shell like(K,L,M,N,O,P,Q)[ascending order]. So,the energy of an electron in K shell should be less than that of an electron in Q shell. But we know that the energy of an electron is constant and is equal to (4.8*10^-10) esu. Then how can this is possible that an electron with a certain energy does not carry any energy while orbiting and it gains the same energy of its orbit while orbiting? Note that there may be some mistakes in my informations as there may be many false informations printed in my book but still I am curious about the question. Thank you!

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    $\begingroup$ Welcome to Chemistry SE. Your question is not very clear. Why do you think electrons don't have energy while orbiting? Also, the energy of an electron is not always constant (it depends on which shell the electron is present in) and it is certainly not $4.8\times10^{-10}esu$. esu is a unit of electric charge so what you have written is the charge on an electron (which is constant - $1.6\times10^{-19}C$ in SI units). $\endgroup$ – kaliaden Apr 17 '13 at 10:18
  • $\begingroup$ Actually I am having confusion.The informations may have some faults. The principle shells such as (K,L,M,N) are called energy shells because they carry energies. If electrons carry certain amount of energy then while orbiting they would loose energy and tend to the nucleus and have a collision with it(according to the general electrodynamics). But the atom is stable. Why? I read my book but it says all the things that I mentioned in the question. How does the electron make the atom stable while orbiting through energized shells (principle shells)and having energies of its own(1.6*10^-19 C)? $\endgroup$ – shiladitya basu Apr 17 '13 at 12:04
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The main confusion I can see in your question is charge vs. energy. Energy is the capacity to do work. Energy comes in many forms. Kinetic energy is the energy of motion. Potential energy is the energy of the system due to its arrangement. Electrons in Bohr shells have energies dependent on the mass of the electron, which is constant, the charge of the electron (constant), the distance from the nucleus (fixed at certain allowable values), and some fundamental constants. Thus, an electron in a shell has unchanging energy. The only way for the electron to change its energy is to move to a different shell. $\Delta E=0$ for an electron in a Bohr shell; $E\neq0$.

An electron in a Bohr shell certainly has kinetic energy because it has orbital angular motion. An electron in a Bohr shell certainly has potential energy because it is being held apart from the oppositely charged nucleus. The total energy of the electron in a Bohr shell ($E_n$) is a function of the principal quantum number, $n$. $Z$ is the number of protons (1 for hydrogen, the only atom for which the Bohr model works) and $R_E$ is the Rydberg energy (related to the Rydberg constant by $R_H=R_E/hc$).

$$E_n=-\frac{Z^2 R_E}{n^2}$$

The Rydberg energy is the smash up of some other more fundamental constants, like the mass of an electron ($m_e = 9.109\times10^{-34} \text{ kg}$), the fundamental unit of charge ($e=1.602\times 10^{-19} \ \text{C}$), Coulomb's constant ($k_e = 8.987 \times 10^9 \ \ce{N m^2 C^{-2}}$ also $k_e=(4\pi\varepsilon_0)^{-1}$, where $\varepsilon_0=8.854\times 10^{-12} \ \text{F/m}$ is the permittivity of a vacuum), and the Bohr radius ($a_0 = 5.29\times 10^{-11} \ \text{m}$). The Bohr radius is itself equal to a mashup of other constants $a_0=\hbar/(m_e c \alpha)$, where $c=2.998\times10^8 \ \text{m/s}$ is the speed of light in a vacuum, $\hbar=1.054\times10^{-34} \ \text{J s}$ is the reduced Planck constant, and $\alpha=7.297\times^{-3}$ is the fine-structure constant. Eventually, $R_E$ simplifies down to a relativistic energy:

$$R_E = \frac{k_e e^2}{2a_o}= \frac{1}{2}m_e c^2 \alpha$$

Electric charge is something else. It is a unit of electrostatic interaction with other charged particles. The electric charge of an object influences the energy of that object in the presence of an electric field, see Coulomb's Law. The (magnitude of) charge on an electron ($e$) is part of the Rydberg energy, so it influences the total energy of the electron.

However, charge$\ \neq\ $energy.

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