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Bohr’s Atom Model Bohr’s Atom Model
Bohr’s Postulates Bohr’s Postulates
- PermissiblePermissible orbitsorbits forfor whichwhich thethe angularangular momentum of the electron is an integral multiple momentum of the electron is an integral multiple of of Where h is Planck’s constant = 6.64 x 10 Where h is Planck’s constant = 6.64 x 10 -34- J-secJ-sec
- An atom radiates energy only when an electron 2. An atom radiates energy only when an electron jumps from a stationary orbit of higher energy( E jumps from a stationary orbit of higher energy( Eii)) to one of the lower energy (E to one of the lower energy (Eff).). Emitted photonEmitted photon frequency is frequency is 2 h
h
E E
i f
The Bohr Formulae The Bohr Formulae
Since mass (M) of the nucleus is much greater Since mass (M) of the nucleus is much greater than the mass (m) of electron, therefore nucleus than the mass (m) of electron, therefore nucleus may be consider as stationary. Thus, may be consider as stationary. Thus, The electrostatic force of attraction between the The electrostatic force of attraction between the nucleus and the electron nucleus and the electron = = (1)(1) and and the centripetal force on the electron = the centripetal force on the electron = (2)(2) 2
r
Ze e
o
r mv 2
Substituting the value of in Eq. (3), we get (6) Therefore, the radius of n th permissible orbit of hydrogen atom ( ) is (7) The radius of the first orbit for hydrogen atom, also known as Bohr radius is r 1 = 0.53 Å 2 v Ze m n h r o 2 2 2 n r n r
e m
n h
o 2 2 2
Calculation of total energy Calculation of total energy
The total energy of the electron on any orbit is the sum of its kinetic and potential energies. Potential energy of the electron is (6)
r
Ze
dr
r
Ze
o r o
2 2 2
Bohr’s interpretation of the Bohr’s interpretation of the Hydrogen spectrum Hydrogen spectrum 2 2 2 1 2 3 4 1 1 8 2 1 h n n me h E E o n n The frequency of the emitted radiation is = In wave number form 2 2 2 1 2 3 4 1 1 8 1 2 1 ch n n me ch E E c o n n
or where = 1.091x 10 7 known as Rydberg constant. 2 2 2 1 1 1 n n R 2 3 4 8 ch me R o
The first line of the Balmer series (n=3) is called the and second (n=4) the Line and so on.
- Paschen series ; n = 4, 5 ,6 …..
- Brackett series ; n = 5, 6, 7….. H H 2 2 1 3 1 n
R
2 2 1 4 1 n
R
- Pfund series ; n = 6, 7, 8 ….. By putting n = ∞ in each one of the series, we get the wavenumber of the series limit. 2 2 1 5 1 n R
13 Using centre of mass theory Mr 1 = mr 2 The angular momentum of the atom about the c. m. Is Where r = r 1
- r 2 and is known as reduced mass of the electron. 2 2 2 1 L Mr mr 2 r M m Mm
Therefore all the equations are identical except that mass m has been replaced by reduced mass Rydberg constant with reduced mass Where (^) M m Rz R 1 1 3 4 8 ch me R o
From spectroscopic data R He = 10972240 m
- and R H = 10967770 m - Therefore, This value is in excellent agreement with the value obtained by other methods. 1837 1 H M m
- The discovery of deuterium Consideration of reduced mass played an important role in the discovery of deuterium. since M D = 2 M H , therefore the spectral lines of deuterium are all shifted slightly to wavelengths shorter than the corresponding ones of ordinary hydrogen. Thus the H α line of deuterium, which arises from a transition from the n = 3 to the n =2 energy level energy level, occurs at a wavelength of 656.1 nm, whereas the H α line of hydrogen occurs at 656.3 nm. This difference in wavelength was responsible for the identification of deuterium in 1932 by H. Urey.
Magnetic moment and angular Magnetic moment and angular momentum momentum Orbital motion of electron in an elliptical orbit of area A with a period of T is equivalent to a current i = e/T in a loop of area A. Applying Ampere’s theorem, this current gives rise to a magnetic dipole moment μ l and given by μ l = iA = eA/T (1) Since the areal velocity in a central orbit is dt d r 2 2 1
Therefore , the area Angular momentum of the electron or (2) from Eq. (1) and (2), dt dt d A r T
0 2 2 1 const dt d L mr 2 m LT dt m L A T 2 2 0 ^
L m e l 2
