Download 2.8 Atomic Spectra of Hydrogen For some time scientist had known

2.8 Atomic Spectra of Hydrogen
For some time scientist had known that every atom, when subjected to high temperatures or an
electrical discharge, emits electromagnetic radiation of characteristic frequencies or each atom
has a characteristic emission spectrum. Because the emission spectra of atoms consists of only
certain discrete frequencies, they are called line spectra. Hydrogen, the lightest and simplest
atom, has the simplest spectrum.
Since atomic spectra are characteristic of the atoms involved, it is reasonable to suspect that the
spectrum depends on the electron distribution in the atom. A detailed analysis of the hydrogen
atomic spectrum turned out to be a major step in the elucidation of the electronic structure of
atoms.
For many years, scientists had tried to find a pattern in the wavelengths or frequencies of the
lines in the hydrogen atomic spectrum. Finally in 1885 the Swiss scientist, Johann Balmer,
showed that,
̅𝜈 = 10973731.6 (1/22 – 1/n2) m-1
when n = 3,4,5….
Where ̅𝜈 = 1/λ = ν/c. This equation is called Balmer formula.
Subsequent to the discovery of the Balmer series of lines, it was found that many other spectral
lines are also present in the non visible regions of the electromagnetic spectrum. Thus Lyman
series originates by the movement of electrons from higher energy states (n = 2,3,4 …..) to the
first energy level for which n =1. Similarly, the Balmer series is accounted by the movement of
electrons from n = 3,4,5,6…… to n =2. Hydrogen atom has a series of spectral lines ; the Lyman
series in U V region, the Balmer series in the visible region, Paschen, Brackett and Pfund series
in the I. R region. By the application of Balmer`s equation, the Swedish Physicist Johannes
Rydberg showed that every line in the entire spectrum of hydrogen can be accounted by the
general expression
1/λ = 𝜐̅ = R [1/n12- 1/n22], Where n1 and n2 represent integers with n2 > n1.
R is the Rydberg constant equal to 10973731.6 m-1 for H atom and 𝜐̅ is the wave number.
Fig.2.2
Spectral series for hydrogen atom
It can be seen that the difference of energy values of electrons between successive orbits become
smaller and smaller and ultimately tend to merge. The spectrum also shows that the line become
closer together at high energy values. The higher energy end is that end where frequency is
increasing or wave length is decreasing.
2.9 Bohr`s Theory of the Atom
In 1913 the Danish physicist Niels Bohr presented a theory of the hydrogen atom that
gave a beautifully simple explanation of the hydrogen spectrum. Bohr made his
contribution by arbitrarily welding Planck`s quantum hypothesis to classical
mechanics.
1. The electron in an atom can exist only in certain states and in these states they
never radiate energy even if it has acceleration. This is the stationary state. Each
stationary state has a definite amount of energy.
2. Energy is emitted only when an electron makes a transition between two
stationary states and not continuously
3. The frequency of the radiation emitted when the transition occurs between two
orbits that differ in energy ΔE
4. The angular momentum of the orbital electron can have only certain values and
that are integral multiple of h/2п
5. The dynamical equilibrium that governs the stationary orbit is determined by
classical mechanics, by balancing the electrostatic attraction against the
centrifugal effect of the orbital motion.
Bohr pictured the electron in a hydrogen atom moving in a circular orbit about the
nucleus.
Fig.2.3
Bohr atom model
There is a nucleus charged Ze and an electron having mass m and charge e
revolving at a distance r from the nucleus, with a velocity v.
The electrostatic force of attraction between the nucleus and the electron =
Ze2/4пε0r2
Where ε0 is permittivity of vacuum, ε0 = 8.85418 x 10-12 C2 N-1 m-2
The centripetal force acquired by the electron = mv2/r
In the circular orbit,
mv2/r = Ze2/4пε0r2
v=
Ze2/4пε0r2.mvr
According to Bohr mvr = nh/2п ,
n = 1,2,3….
So v = Ze2/2ε0nh
This is the velocity of an electron in a circular orbit
The radius of the orbit r can also be calculated. r = nh/2п.mv
r= n2h2εo/Zmпe2
i.e r ∞ n2
Thus the radii of the allowed orbits or Bohr orbits are quantised. The electron can move
around the nucleus in circular orbits only with radii given by the above eqn. The orbit with the
smallest radius is one with n =1 for which
r = (8.8541 x 10-12C2N-1m-2) (6.626 x 10-34Js)2
3.14 (9.109 x 10-31Kg) (1.6022 x 10-19C)2
= 5.292 x 10-11m = 0.529A0
The radius of the first Bohr orbit is often denoted by a
In general we can write, rn = 0.529 x n2/z A0
Problem
Calculate the radius ratio for 2nd orbit of He+ ion & 3rd orbit of Be+++ ion.
r1 (radius of 2nd orbit of He+ ion) = 0.529 (22/2) A0
r2 (radius of 3rd orbit of Be+++ ion) = 0.529 (32/4) A0
r1
r2
=
0.529 (22/2) A0
0.529 (32/4) A0
= 8/9
Energy of Electron in Hydrogen and Hydrogen-like Atoms
Total energy of an electron = Kinetic energy + Potential energy
mv2/r = Ze2/4пε0r2
In the circular orbit,
mv2 = Ze2/4пε0r
Kinetic energy = ½ mv2 = Ze2/8пε0r
r
Potential energy = ∫∞ Ze2/4пε0r2
= - Ze2/4пε0.r
Total energy = ½ (Ze2/4пε0r2) - Ze2/4пε0r2
= - Ze2/8пε0r
Putting the value of r
E = - Ze2.Zmпe2/8пε0n2h2ε0
= - mZ2e4/8ε02n2h2
The negative sign in this equation indicates that the energy states are bound states. n = 1
corresponds to the state of lowest energy. This energy is called the ground state energy. At
ordinary temperature, hydrogen atom, as well as most other atoms and molecules are found
almost exclusively in their ground electronic states. The states of higher energy are called
excited states and are generally unstable with respect to the ground state. An atom or a molecule
in an excited state will usually relax back to the ground state and give off the energy as
electromagnetic radiation.
Bohr assumed that the observed spectrum of the hydrogen atom is due to transition from one
allowed energy state to another and he predicted that the allowed energy difference is given by
ΔE = hυ = me4Z2/8ε02h2 [1/n12 - 1/n22]
Thus, the simple explanation for the line spectrum is that the atoms are constrained to emit
quanta of only a few specific energies depending on the energy difference between the orbits
involved in the movement of electrons. Energetically excited atoms are not able to emit light of
continuously varying wavelengths and therefore not a continuous spectrum is produced from
atoms.
Merits of Bohr Theory
1. Bohr theory explains well the features of hydrogen spectrum and the spectra of hydrogen
like atoms He+, Li2+ etc
2. It helps to calculate radii of various energy levels
3. It also helps to calculate energy associated with various orbits and the velocity of
electrons in various orbits
The Spectrum of the Hydrogen Atom
Bohr was able to derive a correct expression for the energy levels of hydrogen like atoms (i.e
atoms with one electron). Bohr`s formula for the Rydberg constant was in perfect agreement
with the experimental data. However, Bohr`s theory could not be extended successfully to
helium or other atoms with more than one electron.
According to Bohr, the spectrum arises when the electron in the initial higher energy state of
principal quantum number n2 get transmit to the final lower energy state of principal quantum n1
so that the difference of the energies associated with these levels is emitted as photon of
frequency υ
ΔE = hυ = E2 – E1 = - m Z2 e4/8 ε02 n22 h2 – (- m Z2 e4/8 ε02 n12 h2)
υ = m Z2 e4/8 ε02 h3 (1/n12 – 1/n22)
υ̅ = 1/λ = m Z2 e4/8 ε02c h3 (1/n12 – 1/n22)
since c =υλ
υ̅ = RH (1/n12 – 1/n22)
RH = Rydberg constant = m Z2 e4/8 ε02c h3 = 10973731.6 m-1 for H atom
Using this equation, the wave numbers of photons of the various spectral series can be
calculated. Table 2.1 shows the different spectral series of hydrogen with their n1 and n2 values.
The shortest wavelength line any series occurs when n2 is infinitely large so that 1/n22 is zero.
That is, if n2 = ∞, then 1/n22 = 0
Name of series
Value of n1
Value of n2
Region
Lyman
1
2, 3, 4……
U. V
Balmer
2
3, 4, 5…….
Visible
Paschen
3
4, 5, 6……
I.R
Brackett
4
5, 6, 7…….
I.R
Pfund
5
6, 7, 8……..
I.R
Table 2.1 Different spectral series of hydrogen
Problem
1. Calculate the ionization potential of the hydrogen atom
The ionization potential of an atom or molecule is the energy needed to completely remove an
electron from the atom or molecule in its ground state
E = RH [1/n12 - 1/n22]
n1 =1, n2 = ∞
= 2.179 x 10-11 erg
2. Calculate the value of the Rydberg constant of hydrogen atom in eV.
RH = 10973731.6 m-1 for H atom and 1eV = 8.0655 x 105m-1
RH = 13.6 eV
Fig.2.4 Energy levels for the hydrogen atom as calculated from the Bohr theory
Defects of Bohr model
Though, Bohr model could explain the stability of atoms and the origin of line spectrum, the
model was not satisfactory for many observations.
(1). It could not explain the fine spectrum- The spectrum with additional lines when observed
with a high resolution instrument.
(2). It could not explain the splitting of the spectral lines when the source emitting the spectral
lines were placed in a magnetic field (Zeeman effect) or electric field (Stark effect)
(3). The model could not explain why atoms react
It is to be noted that Sommerfield later modified the Bohr model by adding elliptical orbits in
addition to the circular orbits by which a more appropriate explanation was feasible.