Question

Derive the energy expression for an eletron is the hydrogen atom using Bohr atom model.

Answer 1

Bohr postulates is finds the allowed energies of the atom for different allowed orbits of the electron.

1. The electron in an atom moves around the nucleus in circular orbits under the influence of Coulomb electrostatic force of attraction.
2. Electrons in an atom revolve around the nucleus only in certain discrete orbits called stationary orbits where it does not radiate electromagnetic energy.
   Angular momentum quantization condition,
   L = `"nh" = "nh"/(2pi)`
   h – planck’s constant
   n – principal quantum number of the orbit.
3. An electron can jump from one orbit to another orbit by absorbing or emitting a photon whose energy is equal to the difference in energy (∆E) between the two orbital levels.
   energy quantization condition
   `delta "E" = "E"_"final" - "E"_"initial" = "hv" = "hc"/lambda`
   

   λ – wavelength of the radiation
   c – speed of light
   υ – frequency of the radiation

   Radius of the orbit of the electron:

   
   Electron revolving around the nucleus

1. The nucleus has a positive charge +Ze.
2. Let z be the atomic number of the atom.
3. Let -e be the charge of the electron.
4. The nucleus at rest and an electron revolving around the nucleus in a circular orbit of radius r_n with a constant speed υ_n
   Coulomb’s law, the force of attraction between the nucleus and the electron is
   `vec"F"_"coulomb" = 1/(4 piepsilon_0) ((+ "Ze")(-"e"))/"r"_"n"^2 hat"r"`
   `= - 1/(4 piepsilon_0) ("Ze"^2)/("r"_"n"^2) hat"r"`
5. This force provides necessary centripetal force.
   `vec"F" _"centripetal" = ("mv"_"n"^2)/"r"_"n" hat"r"`
   where m be the mass of the electron that moves with a velocity υ_n in a circular orbit.
   Therfore,
   `1/(4 piepsilon_0) ("Ze"^2)/("r"_"n"^2) = ("mv"_"n"^2)/"r"_"n"`
   `"r"_"n" = (4piepsilon_0 ("mv"_"n""r"_"n")^2)/("Zme"^2)`  ....(1)
   From Bohr's assumption, the angular momentum quantization condition,
   `"mv"_"n""r"_"n" = "l"_"n" = nh`,
   `therefore "r"_"n" (4piepsilon_0("nh")^2)/("Zme"^2) = (4piepsilon_0"n" "h"^2)/("Zme"^2)`
   `"r"_"n" = ((epsilon_0"h"^2)/(pi"me"^2))"n"^2/"Z" (therefore "h" = "h"/(2pi))`
   Where n∈N ε₀, n, e and π → are constant
   radius of the orbit,
   r_n = a₀ `"n"^2/"Z"`
   Bohr radius a₀ = `(epsilon_0"h"^2)/(pi"me"^2)`
   a₀ = 0.529 Å
   Bohr radius is also used as unit of length called Bohr

   1 Bohr = 0.53 Å

   For hydrogen atom Z = 1
   r_n = a₀ n²
   r_n α n²
   For the first orbit (ground sate)
   r₁ = a₀
   r₂ = 4a₀ = 4r₁
   r₃ = 9a₀ = 9r₁
   r₄ = 16a₀ = 16r₁

   The energy of an electron in the nth orbit:
   Since the electrostatic force is a conservative force, the potential energy for the nth orbit is
   `"U"_"n" = 1/(4piepsilon_0) ((+ "Ze")(- "e"))/"r"_"n" = - 1/(4piepsilon_0) "Ze"^2/"r"_"n"`
   `= - 1/(4epsilon_0^2) ("Z"^2"me"^4)/("h"^2"n"^2) (therefore "r"_"n" = (epsilon_0"h"^2)/(pi"me"^2) "n"^2/"Z")`
   The kinetic energy for the n^th orbit 
   `"KE"_"n" = 1/2 "mv"_"n"^2 = "me"^4/(8epsilon_0^2 "h"^2) "Z"^2/"n"^2`
   This implies that `"U"_"n" = - 2 "KE"_"n".`
   Total energy in the nth orbit
   `"E"_"n" = "KE"_"n" + "U"_"n" = "KE"_"n" - 2 "KE"_"n" = - "KE"_"n"`
   `"E"_"n" = (- "me"^4)/(8epsilon_0^2"h"^2) "Z"^2/"n"^2`
   For hydrogen atom (Z = 1),
   `"E"_"n" = (- "me"^4)/(8epsilon_0^2"h"^2) 1/"n"^2`joule
   `"E"_"n" = - 13.6 1/"n"^2 "eV"`

6. The negative sign in the equation indicates that the electron is bound to the nucleus

7. Where n stands for a principal quantum number.

8. The energies of the excited states come closer and closer together when the principal quantum number n takes higher values.

9. The ground state energy of hydrogen (-13.6 eV) is used as a unit of energy called Rydberg (lRydberg = -13.6eV).

10. For the first orbit (ground state), the total energy of the electron is E₁ = -13.6 eV.

11. For the second orbit (first excited state), the total energy of the electron is E₂ = – 3.4eV.

12. For the third orbit (second excited state), the total energy of electron ie E₃ = -1.51 eV and so on.

13. When the electron is taken away to an infinite distance (very far distance) from nucleus, both the potential energy and kinetic energy terms vanish and hence the total energy also vanishes.