The Finite Physical Quantities of the Bohr Model of Hydrogen Atom

 

The Finite Physical Quantities of the Bohr Model of Hydrogen Atom 

Pavle I. Premović

Laboratory for Geochemistry, Cosmochemistry&Astrochemistry,

University of Niš, pavleipremovic@yahoo.com, Niš, Serbia

According to the Bohr model, the hydrogen atom consists an electron which orbits a proton The proton has a positive charge equal in magnitude to a unit of electron charge e (= 1.60 × 10^-19 C) and its radius of about 0.85 × 10^-15 m. The radius of the ground state orbit of hydrogen atom is symbolized as a₀ and it is about 5.3 × 10^-11 m. 

The energy expression for the Bohr model of hydrogen atom is

 

E_n = –2π²m_ee⁴/n²h²

where:

 

m_e = the rest mass of electron

n (= 1, 2, 3,..) is a quantum number

h (= 6.63 × 10^-34 J sec) is Planck’s constant.[1] 

We mark with E₁ the energy of an electron in the ground orbit of hydrogen atom with n = 1. This energy is – 2.18 ×10^-18 J (or – 13.6 eV). The energies of all the following orbits successively with 1, 2, 3,.. by their quantum number n and if n → ∞, E_n → 0.

The Coulomb force of attraction between the proton and electron of hydrogen atom is 

Fp = kee2/Rn2 

where k_e is the Coulomb constant (= 8,988 × 10⁹ N m² C^-2)[2] and R_n is the distance between these particles. This force causes the electron to deflect from a straight line to a circular orbit around the proton. This is centripetal force and it is balanced by a centrifugal force F_f given by the formula m_eυ_n²/R_n where υ_n is the electron speed. In the equation form 

F_p = F_f 

or

 m_eυ_n²/R_n = k_ee²/R_n²    … (1).

The Bohr quantum condition is defined by the following equation 

m_eυ_nR_n = nħ    … (2)[3] 

where ħ (= h/2π = 1.055 × 10⁻³⁴ J sec) is the reduced Planck constant. Combining this equation and eqn. (1) we get, after a bit of algebra,

 υ_n = k_ee²/nħ. 

This equation indicates that υ_n ≤ k_ee²/ħ, so its maximum value υ₁ = k_ee²/ħ. Plugging into it the above known values we find υ₁ ≈ 2.2 × 10⁶ m sec^-1. Substituting k_ee²/ħ of this equation with υ₁, we have

 υ_n = υ₁/n. 

If n → ∞ then υ_n → 0 and the corresponding acceleration a_n → 0. 

A combination of eqns. (2) and (3), after some algebra, also yields 

R_n = (ħ²/k_em_ee²)n²   … (3). 

By setting n = 1 into this equation and after some algebra, we obtain 

R_n = a₀n².  

So, when n → ∞ then E_n → 0, R_n → ∞, υ_n → 0 and a_n → 0. 

When an excited electron from the orbit with the highest energy E_∞ = 0 J returns to the ground state energy orbit it emits a photon with an energy of 13.6 eV. The radius of this highest energy orbit or the distance between an electron in this orbit and the proton is infinite.  A question arises: how long it takes the electron to reach the ground state? Since R_n → ∞ then the time for this transfer t → ∞.

Let us assume that the energy of the highest energy electron orbit is not zero but it has an exceedingly low value, E_i. In this case, we would deal with n_i, R_i and t_i having finite but exceedingly large values. The corresponding speed υ_i and acceleration a_i would have an exceedingly low value. 

When the electron reaches the ground state orbit its speed υ₁ would be about 2.2 × 10⁶ m sec^-1 {1} and its average speed would be υ_av = (υ₁ + υ_i)/2 ≈ υ₁/2 ≈ 1 × 10⁶ m sec^-1.

Premović [1] estimated that an electron situated at a distance of Bohr’s radius from the proton would reach an acceleration of about 10³⁰ m sec^-2 or a₁ ≈ 10³⁰ m sec^-2. Its average acceleration would be a_av = (a₁ + a_i)/2 or a_av ≈ 5 × 10²⁹ m sec^-2. 

Since a_av = υ_av/t_i find that t_i = υ_av/ a_av ≈ 1 × 10⁶ m sec^-1/5 × 10²⁹ m sec^-2 ≈ 2 × 10^-24 sec. This time interval is about 2000 times smaller than the smallest time interval measured to date: 10^-21 sec. 

We can estimate E_i another form of the Heisenberg relations relating to energy and time. In equation form 

E_it_i = h 

or 

 E_i = h/t_i.

Plugging the above value of t_i, we get E_i ≈ 10^-10 J. However, now we face a problem. We know that R_i = υ_av × t_i or R_i ≈ 1 × 10⁶ m sec^-1 × 2 × 10^-24 sec ≈ 2 × 10^-18 m. This is impossible since, as we noted above, R_i has to be exceedingly large. 

Reference 

{1} P. I. Premović, The Coulomb electron speed and acceleration in Bohr’s model of the hydrogen atom. The General Science Journal, April 2023.

[1] Although the Bohr model of the hydrogen atom has many shortcomings, the energy equation of the Schroedinger quantum mechanical model for this atom gave the same energy equation as the Bohr equation. In both the Bohr model and quantum mechanics, the energy is proportional to 1/n², according to eqn. (1).

[2] To avoid confusion in further text, the SI units are given in italics. 

[3] We know that m_eυ_nR is the angular momentum of an electron in its orbit which is, according to Bohr’s model, quantized.