## Bohr Model of Hydrogen Energy Levels

**Bohr Model of hydrogen atom** was adopted by Neils Bohr in 1913 for the explanation of Rutherford theory and the atomic spectrum of hydrogen energy levels. According to classical mechanics, when a charged electron is subjected to acceleration, it emits radiation and energy released to hit the nucleus of an atom. Therefore, according to Rutherford atomic theory, the electron loses energy continuously and the observed spectra should be continuous spectra. But the actual electromagnetic spectrum of the hydrogen atom consists of well-defined lines with definite frequencies and wavelengths.

For an explanation of the atomic spectrum, Bohr adopted the Rutherford nuclear model for the hydrogen atom or one electronic system in which an electron revolves around the proton of the nucleus. The Bohr atomic model of hydrogen atom alike with the solar system of our planet is based on several postulates which proved by wave mechanical consideration in chemistry or physics.

### Bohr Model of Hydrogen Atom Postulates

An atom or hydrogen atom possesses several stable circular orbitals in which an electron can stay. Therefore an electron stays in a particular orbit where no emission or absorption of energy occurs. These orbits are called the energy levels of an atom. The energy levels designated by numbers 1, 2, 3, …….. or capital letters, K, L, M, …….. starting from the nucleus of the atom. Therefore the energy associated with a certain hydrogen energy levels increases with the increase of its distance from the nucleus. If E_{1}, E_{2}, E_{3} …… denotes the number of energy levels of 1 (K-shell), 2 (L-shell), 3 (M-shell) ……. then the order of energy levels are, E_{1}ㄑE_{2}ㄑE_{3}ㄑ……

When an electron can jump from one orbit to another higher energy on the absorption of unit energy but one orbit to another lower energy with the emission of unit energy.

The angular momentum of an electron moving in an orbit is an integral multiple of h/2π. This integral multiple is known as the principal quantum energy levels of the hydrogen atom. Therefore, mvr = nh/2π, where m = mass of an electron, v = tangential velocity of an electron, r = radius of Bohr energy levels.

### Nucleus Electrons Attraction in Hydrogen Atom

Let the nucleus has a mass m′ and the electron has mass m, the radius of the circular orbit = r, and the linear velocity of the electron = v.

On the revolving electron in the hydrogen atom, two types of forces acting, centrifugal force and the electric force of attraction. Hence the centrifugal force = mv^{2}/r and electric force of attraction between two opposite charges = e^{2}/r^{2} (given from Coulomb’s law for the polar bond).

Centrifugal force and electric force acted in the opposite direction. Thus the electron may keep on revolving in hydrogen atom energy levels if these two forces act in the opposite direction must balancing by each other. Therefore, mv^{2}/r = e^{2}/r.

### Atomic Radius of Bohr Energy Levels

The remarkable study from Niels Bohr atomic theory, the angular momentum of the electron is an integral multiple of h/2π. Therefore, mvr = nh/2π, where mvr = angular momentum and n = principal quantum number having the values 1,2,3, ……∞.

If we putting the value of v on the Bohr force equation, the radius (r) = n^{2}h^{2}/4π^{2}me^{2}.

### First Orbit of Hydrogen Atom Radius

From the above equation, we can easily find out the radius of the permitted energy levels in terms of the quantum number of the hydrogen atom. When n = 1, the radius of the first stationary orbit of hydrogen (r_{1}) = 0.529 × 10^{-8} cm = 0.529 Å. Therefore, the ratio of first Bohr orbit and nth orbit of the hydrogen atom, r_{n} = n^{2} × r_{1}.

Question: Calculate the radius of the second orbit of hydrogen atom if the radius of the first orbit of hydrogen = 0.529 Å.

Answer: The radius of the second orbit of the hydrogen atom (r_{2}) = n^{2} × r_{1} = 2.12 Å

### Velocity of an Electron in Hydrogen Atom

From the principle of quantization to the revolving electron in the hydrogen atom, mvr = nh/2π. When we putting the values of r in mvr = nh/2π and solving, we have velocity (v) in the Bohr model = 2πe^{2}/nh. Therefore, the velocity of the second orbit will be one half of the first orbit and one-third of the first orbit, and so on.

Problem: Calculate the velocity of the hydrogen electron in the first and third energy levels of the Bohr model. How to calculate the number of rotation of an electron per second in the third energy level in the Bohr model?

Solution: The velocity of an electron nth level of the hydrogen atom (v_{n}) = 2πe^{2}/nh, where n = 1, 2, 3, …….. Thus the velocity of an electron in the first energy level (v_{1}) = 2πe^{2}/1^{2} × h = 2πe^{2}/h = 2.188 × 10^{8} cm sec^{-1}.

The velocity of the third energy level (v_{3}) = v_{1}/3 = (2.188 × 10^{8} cm sec^{-1})/3 = 7.30 × 10^{7} cm sec^{-1}. The radius of the third energy level = 3^{2} × 0.529 × 10^{-8} cm. Hence, the circumference of the third energy level = 2πr = 2 × 3.14 × 0.529 × 10^{-8} cm. Therefore, the rotation of an electron per second in the third energy level = (7.30 × 10^{7})/(2 × 3.14 × 9 × 0.529 × 10^{-8} ) = 2.44 × 10^{14} sec^{-1}.

### Energy of an Electron in Hydrogen Atom

The energy of an electron moving in one particular energy level of the hydrogen atom in the Bohr models can be calculated by the total energy or the sum of the kinetic energy and the potential energy of an electron in physics and chemical science. Here kinetic energy = ½ mv^{2} and potential energy for moving electron = e^{2}/r. The total energy for the nth level of the hydrogen atom in Bohr model, E_{n} = (1/2mv^{2}) + (- e^{2}/r) = – 1/2mv^{2}. Therefore, E_{n} = – e^{2}/2r, where mv^{2} = e^{2}/r. When we putting the value of r in the above equation, the total energy in the Bohr model, E_{total} = – 2π^{2}me^{4}/n^{2}h^{2}.

In learning chemistry, E_{n} = E_{1}/n^{2}, where E_{1} = energy of the first orbit of the hydrogen atom in the Bohr model. As the principal quantum number (n) increases the energy becomes less negative and hence the Bohr system becomes less stable. Because with increasing n, r also increases and orbit makes less stable. If the energies associated with 1st, 2nd, 3rd,…., nth orbits in the Bohr model are E_{1}, E_{2}, E_{3} … E_{n}, hence, E_{1}ㄑE_{2}ㄑE_{3}ㄑ……..ㄑE_{n}.

### Electron Energy Calculation in Bohr Model

The energy of the moving electron in the first energy levels obtained by putting n=1 in the Bohr model of the hydrogen atom. Therefore, the energy of first orbit in Bohr model = – 21.79 × 10^{-12} erg = – 13.6 eV = – 21.79 × 10^{-19} Joule = – 313.6 Kcal.

Question: H, H^{+}, He^{+,} and Li^{+2} – for which of the species Bohr’s model of the hydrogen atom is not applicable?

Answer: From the above species H, He^{+}, Li^{+2} contain one electron but H^{+}-ion has no electron. The Bohr’s model of hydrogen atom applicable to the one electronic system in physics or chemistry but for H^{+}-ion Bohr atomic model is not applicable.