Electron Behavior and Periodic Properties of Elements

The Bohr Atom

Niels Bohr developed a model of the atom in which electrons have quantized energies; that is, they can only be found at specific distances from the nucleus.
To explain why electrons seem to be associated with discrete energies, in 1913 Danish physicist Niels Bohr proposed the Bohr model of the atom. In this model, electrons are found only in shells located at a specific distance from the nucleus. This model is still in common use today, but it only works for hydrogen-like atoms/ions, which are one-electron systems. The shells are numbered beginning with the shell closest to the nucleus and increasing in number farther away from the nucleus.

Bohr Model of an Atom

The Bohr model of the atom, developed by Danish physicist Niels Bohr, claims that electrons are found only in shells located at specific distances from the nucleus.
Bohr calculated the energy in electron volts (eV) of the electrons in the nth energy level for hydrogen as
E=1n2×13.6eVE=-\frac1{n^2}\times13.6\;\rm{eV}
Thus, the energy of an electron in the n=1n=1 orbital is 112×13.6eV-\frac1{1^2}\times13.6\;\rm{eV}, or -13.6 eV.

This energy is always negative because it is in relation to a free electron separated from its nucleus. An electron in orbit around a nucleus is much more stable than a free electron, so it has a lower energy.

Bohr proposed that an atom that has no excited electrons has all its electrons in their ground state—the electron state that is lowest in energy. When energy is added, such as when light strikes the electrons, they become excited and enter an excited state—an electron state that has a higher energy level than the ground state. When this happens the electrons jump from a lower orbital to a higher one. However, they cannot maintain their excited state because they are unstable, or do not have the energy to maintain their location. So they drop back down to the ground state, emitting energy in the form of light or heat. Only light of a certain energy can cause an electron to jump from its ground state to an excited state.
When light strikes an electron, it provides excitation energy, causing the electron to jump to orbitals in a higher energy state. The excited state is unstable, so the electron relaxes back to its ground state, emitting light in the process.
The energy of this transition (ΔE\Delta{E}) can be calculated by subtracting the energy of the electron in the ground state from the energy of the electron in the excited state.
ΔE=EexcitedEground=(1(nexcited)2×13.6eV)(1(nground)2×13.6eV)=(1(nground)21(nexcited)2)×13.6eV\begin{aligned}\Delta E&=E_{\rm{excited}}-E_{\rm{ground}}\\&=\left(-\frac{1}{\left({n_{\rm {excited}}}\right)^2}\times13.6\;\rm{eV}\right)-\left(-\frac {1}{\left({n_{\rm {ground}}}\right)^2}\times13.6\;\rm{eV}\right)\\ &= \left(\frac1{\left({n_{\rm {ground}}}\right)^2}-\frac1{\left({n_{\rm {excited}}}\right)^2}\right)\times13.6\;\rm{eV}\end{aligned}

Electron Transitions for a Hydrogen Atom

A hydrogen atom has four electron transition states, each subdivided according to the excited state from where an electron is originally located. The light emitted by an electron depends on the energy difference between the excited and ground state. Notice that emissions from the Balmer series explain the four line colors of hydrogen's emission spectrum.
Planck's constant (h), 6.62607×1034Js6.62607\times 10^{-34}\;\rm{J} \cdot \rm{s}, describes the ratio of energy to frequency of a photon, so it and the frequency (ν\nu) of the light absorbed or emitted during the change can be used to solve for ΔE\Delta{E} instead.
ΔE=hν\Delta E=h\nu
Because frequency is calculated by dividing speed by wavelength, and the speed of light is a known constant, the equation can be rewritten using the speed of light (c), 3.00×108m/s3.00\times10^8\;\rm{ m/s}, and wavelength (λ\lambda).
ΔE=hcλ\Delta E=\frac {hc}\lambda
Thus, for an electron giving off light at 656 nm, or 6.56×107m6.56\times10^{-7}\,\rm{m}, the change in energy of this transition of electron from the excited to ground state can be calculated.
ΔE=(6.62607×1034Js)(3.00×108m/s)6.56×107m=3.03×1019J\begin{aligned}\Delta E&=\frac{(6.62607\times10^{-34}\;\rm J\cdot\rm s) (3.00\times10^8\;\rm m/ s)}{6.56\times10^{-7}\;\rm{m}}\\&=3.03\times10^{-19}\;\rm J\end{aligned}
For hydrogen the Balmer series relates the wavelength of light emitted by the transition from excited to ground electron state to the electron transition. The Balmer formula is given as
1λ=RH(1nf21ni2)\frac1\lambda=R_{\rm{H}} \left( \frac{1}{{n_f}^2}-\frac{1}{{n_i}^2} \right)
where λ\lambda is wavelength, RH is the Rydberg constant for hydrogen (1.097×107m11.097\times10^7\;\rm{m}^{-1}), nf is the final value of n (which is 2 for hydrogen), and ni is the initial value of n, which must be an integer greater than 2. For example, for the transition from ni of 3:
1λ=RH(1nf21ni2)=(1.097×107m1)(122132)=16.56×107mλ=656nm\begin{aligned}\frac{1}{\lambda}&=R_{\rm{H}}\left(\frac{1}{{n_f}^2}-\frac{1}{{n_i}^2}\right)\\&=(1.097\times10^7\;\rm{m}^{-1})\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\\&=\frac{1}{6.56\times10^{-7}\,\rm{m}}\\{\lambda}&={656\,\rm{nm}}\end{aligned}
Notably, the observed emission spectrum of hydrogen precisely aligns with values calculated using this formula. This formula gives very similar information to that of the energy of the transition, ΔE\Delta{E}, the difference being the units—nm (the wavelength of light) for the Balmer formula and eV (the energy of the light) for ΔE\Delta{E}.