Bohr Model of the Hydrogen Atom

Bohr Model

The main features of Rutherford's model, viz. the nucleus and the electrons orbiting it under the action of the Coulomb's law of electrostatic attraction, were retained in Bohr's theory. In addition Bohr introduced the concept of 'radiation less orbits' or 'stationary states' in which the electron revolves around the nucleus, but does not radiate contrary to the laws of electromagnetism. This was a hypothesis, but at least a working one.

Radiation occurred only when an electron made a transition from one stationary state to another. The difference between the energies of the two states was radiated as a single photon. Absorption occurred when a transition occurred from a lower stationary state to a higher stationary state.

The third principle invoked by Bohr was the correspondence principle: i.e. the spectrum is continuous and the frequency of light emitted equals the frequency of the electron. In the classical limit.

Using these general arguments and the existing body of knowledge, Bohr postulated the following:

(i) The electron in an atom has only certain definite stationary states of motion allowed to it, called as energy levels. Each energy level has a definite energy associated with it. In each of these energy levels, electrons move in circular orbit around the positive nucleus. The necessary centripetal force is provided by the electrostatic attraction of the protons in the nucleus. As one moves away from the nucleus, the energy of the states increases.

(ii) These states of allowed electronic motion are those in which the angular momentum of an electron is an integral multiple of h/2$\pi$ or one can say that the angular momentum of an electron is quantised.

$\Rightarrow$ Angular momentum = m v r = n (h/2$\pi$)

Where m is the mass, v is the velocity, r is the radius of the orbit, h is Planck's constant and n is a positive integer.

(iii) When an atom is in one of these states, it does not radiate any energy but whenever there is a transition from one state to other, energy is emitted or absorbed depending upon the nature of transition.

(iv) When an electron jumps from higher energy state to the lower energy state, it emits radiations in form of photons or quanta. However, when an electron moves from lower energy state to a higher state, energy is absorbed, again in form of photons.

(v) The energy of a photon emitted or absorbed is given by using Planck's relation

(E = hv). If E₁ be the energy of any lower energy state and E₂ be the energy of any higher energy state then the energy of the photon (emitted or absorbed) is given as $\Delta$E:

$\Delta$E = E₂ – E₁ = h v

Where h = Planck's constant and v = frequency of radiation emitted or absorbed.

Note: Bohr's Model is applicable only to one-electron atoms like: He⁺, Li⁺⁺ apart from H-atom.

In the following section we will define the radius of orbits around the nucleus (Bohr's orbits), the velocity and energy of an electron in various orbits around the nucleus.

We will take

r_n : radius of nth orbit

v_n : velocity of electron in nth state (orbit)

E_n : energy of nth state

m : mass of an electron (9.1 x 10^–31 Kg)

Z : atomic number (No. of Protons)

K = 1/(4$\pi$$\epsilon$₀) = constant = 9 x 10⁹ N m² C^–2

h : Planck's constant (6.67 x 10^–34 J-s)

c : velocity of light (3 x 10⁸ m/s)

R : Rydberg constant (1.097 x 10⁷ m^–1)

e : Charge on an electron (1.6 x 10^–19 C)

v : frequency of the radiation emitted or absorbed

$\overline{v}$ : wave number of the spectral line in the atomic spectra

From first postulate

$\frac{m{v}^2}{r_n}=\frac{KZ{e}^2}{r_n^2}$

From second postulate

$m{v}_n{r}_n=\frac{nh}{2\pi }$

Solving for $r_n$ and $v_n$, we have:

$radius=r_n=\frac{n^2{h}^2}{4{\pi }^{2}Km{e}^{2}Z}$

$=0.53\times {10}^{-10}\frac{n^2}{Z}m$

$=0.53\frac{n^2}{Z}A$

$velocity=v_n=\frac{2\pi KZ{e}^2}{nh}$

$=2.165\times {10}^6\frac{Z}{n}m/s$

$\bullet$ The energy of an electron in nth state E_n is given by:

E_n = KE + PE

$E_n=\frac{1}{2}m{v}^2+\left(-\frac{KZ{e}^2}{r_n}\right)=-\frac{1}{2}\frac{KZ{e}^2}{r_n}$

$\Rightarrow E_n=\frac{-2{\pi }^2{K}^2{e}^{4}m{Z}^2}{n^2{h}^2}\left(\text{putting vaue of }{\text{r}}_{\text{n}}\right)$

$=-2.178\times {10}^{-18}\frac{Z^2}{n^2}J/atom$

$=-13.6\frac{Z^2}{n^2}eV/atom$

$=-1312\frac{Z^2}{n^2}kJ/mol$

$\bullet$ When an electron jumps from an outer orbit (higher energy) n₂ to an inner orbit (lower energy) n₁, then the energy emitted in form of radiation is given by:

$\Delta E=E_{n_2}=E_{n_1}=\frac{2{\pi }^2{e}^{4}m{Z}^2{K}^2}{h^2}\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$

As we know that E = hv, c = v$\lambda$ and $\bar{v}=1/\lambda$

$\Rightarrow \bar{v}=\frac{\Delta E}{hc}$

$\Rightarrow \bar{v}=\frac{1}{\lambda }=\frac{2{\pi }^2{e}^{4}m{Z}^2{K}^2}{c{h}^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Now this can be represented as:

$\bar{v}=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Where $R=\frac{2{\pi }^2{e}^4{K}^{2}m}{c{h}^3}$

Illustration : Radiation emitted in the transition n = 3 to n = 2 orbit in a hydrogen atom falls on a metal to produce photoelectrons, the electrons emitted from the metal surface with the maximum kinetic energy are allowed to pass through a magnetic field of strength 3.125x10^–4 T and it is observed that that the paths haves radius of curvature 10 mm

Find (a) the kinetic energy of the fastest photo electrons.

(b) the work function of the metal.

(c) the wavelength of the radiation.

Solution: The kinetic energy of the fastest photoelectrons is given by

$\Delta E=h{v}_{23}=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)eV$

The Kinetic energy of the fastest photoelectrons is given by

$\frac{m{v}^2}{r}=Bev$

or mv = p = Ber

$\therefore$ Kinetic Energy = $\frac{1}{2}m{v}^2=\frac{B^2{e}^2{r}^2}{2m}$

= $\frac{{\left(3.125\times {10}^{-4}\right)}^2\times \left(1.6\times {10}^{-19}\right)\times {10}^{-4}}{2\times 9\times {10}^{-31}}J$

= 0.86 eV

(a) The kinetic energy of the fastest electron is 0.86 eV

(b) The work function, $\varphi$ is given by

$\varphi$ = (1.9 – 0.86) eV = 1.04 eV

(c) The wavelength of the emitted radiation is,

$\lambda =\frac{hc}{E}=\frac{\left(6.63\times {10}^{-34}\right)\times \left(3\times {10}^{-8}\right)}{1.3\times 1.6\times {10}^{-19}}m$

= 6.54 × 10^{–7 m}

= 6540Å