B.sc 1st year Book (Page 1)
Bohr Atomic model
What is Bohr’s Atomic Model?
In 1913, Niels Bohr’s proposed a theory for the hydrogen atom which is based on the quantum theory that some physical quantities only take discrete values. In an atom, The electrons(e^{–}) revolve around the positively charged nucleus(+) in a defined circular path called shells or orbits. Each shell or Orbit has an examined fixed energy and these circular shell or orbits are called orbitals shell.
Bohr Atomic model :
Bohr’s postulates :
1. The electron revolves in circular orbits around the nucleus only in certain allowed energy states called stationary states or energy levels (as the earth revolves around the sun).
quantized and is an integral multiple of
$$
\frac{h}{2 \pi}
$$
$$
m v r=\frac{n h}{2 \pi}
$$
Related Topic | Atomic Structure and Periodic Table |
Bohr’s Theory of Hydrogen and Hydrogen-like atoms (e.g. He*, Ll^{2+}, Be” etc.)
electrostatic force of attraction between the electron and nucleus is balanced by Centrifugal force which results in the throw of the electron occurring out of the orbit. It is the mass of the electron moving with velocity in a circular orbit of radius as shown in figure 1.01.
Since centrifugal force = coulombic force for attraction
$$ \frac{mv^2}{I}\:=\:\frac{Ze^2}{r^2} $$
Hence,
$$ v^2\:=\:\frac{Ze^2}{mr}$$
According to Bohr’s second postulate
Or $$ mvr=\:n\frac{h}{2\pi } $$
Or $$ v^2=\:\frac{n^2h^2}{4\pi ^2m^2r^2} $$
On comparing the equation (v) and (vi) we get
$$ \frac{2e^2}{mr}=\:\frac{n^2h^2}{4\pi ^2m^2r^2} $$
$$ n_n=\:\frac{n^2h^2}{4\pi ^2m^2e^2}\:\left(vii\right) $$
For hydrogen atom n =1 and Z = 1
Hence
$$ r=\:\frac{h^2}{4\pi ^2m^2e^2}}\:\left(viii\right) $$
Substituting the values h = 6.627 x 10-27 erg sec, m = 9.1 x 10-28g and e = 4.80 x 10-10 e.s.u in equation (viii) we have
r = 0.529 A0
Thus, atomic radius of the H-atom is 0.529 A0.
Calculation of total energy of an electron: The total energy of an electron E is given by –
E = K.E + P.E
Or
$$ E\:=\:\frac{1}{2}mv^2\:-\:\frac{Ze^2}{r} $$
( $$ K.\:E\:=\:\frac{1}{2}mv^2\:since\:and\:P.E\:=\:-\:\frac{Ze^2}{r}\: $$ )
From Equation (v)
$$ \:\frac{1}{2}mv^2\:=\:-\:\frac{2e^2}{2r}\:…..\:\left(x\right) $$
$$ E\:=\:-\:\frac{Ze^2}{2r}\: $$
Now putting the value of r from equation (vii) we get
$$ E\:=\:\frac{Ze^24\pi ^2mZe^2}{2n^2h^2}\: $$
$$ E\:=\:\frac{Ze^24\pi ^2mZ^{ }e^2}{2n^2h^2} $$
$$ E\:=\:\frac{2\pi ^2mZ^2e^4}{n^2h^2}\:….\:\left(xii\right) $$
In the above equation for a particular atom all the quantities except n’ are constant. Therefore
$$ E\:\infty \:-\:\frac{1}{n^2}\:….\:\left(xiii\right) $$
Since there is a negative sign, the value of ‘E’ increases as ‘n’ increases.
Explanation of Hydrogen spectrum on the basis of Bohr’s theory If electron. jumps from one energy level to another then,
$$ E\:_1\:=\:\frac{2\pi ^2me^4}{n\frac{2}{1}h^2}\:and\:E_2=\:\frac{2\pi me^4}{n\frac{2}{2}h^2}\:\:\:\left(Z=1\right) $$
Therefore,
$$ E_2\:-\:E\:_1\:=\:-\frac{2\pi \:me^4}{n\frac{2}{2}h^2}+\frac{2\pi ^2me^4}{n\frac{2}{1}h^2}\: $$
Or $$ E_2\:-\:E\:_1\:=\:-\frac{2\pi \:me^4}{n^2}\left(\frac{1}{n\frac{2}{1}}\:-\frac{1}{n\frac{2}{2}}\right) $$
We know that E2 – E1 = hv Hence,
$$ hv\:=\:-\frac{2\pi \:me^4}{n^2}\left(\frac{1}{n\frac{2}{1}}\:-\frac{1}{n\frac{2}{2}}\right) $$
Or $$ v\:=\:-\frac{2\pi \:me^4}{n^2}\left(\frac{1}{n\frac{2}{1}}\:-\frac{1}{n\frac{2}{2}}\right)\left(xv\right) $$
$$ Since\:c\:=\:cλ\:or\:v=\frac{c}{λ}=cv\::\:where\:’c’\:is\:velocity\:of\:light\:and\:λ\:is\:the\:wavelength.\:\: $$
Since
Or $$ \overline{v}\:=\:\frac{2\pi ^2me^4}{eh^3}\left(\frac{1}{n\frac{2}{1}}-\frac{1}{n\frac{2}{2}}\right)….\:\left(xvi\right) $$
On comparing equation (xvi) with Rydberg’s empirical equation, we get Rydberg’s empirical equation $$ \overline{v}\:=R\left(\frac{1}{n\frac{2}{1}}-\frac{1}{n\frac{2}{2}}\right) $$ (R = Rydberg constant)
I.e.
$$ \overline{v}\:=R\left(\frac{1}{n\frac{2}{1}}-\frac{1}{n\frac{2}{2}}\right)\:=\:\frac{2\pi ^2me^4}{h^3}\left(\frac{1}{n\frac{2}{1}}-\frac{1}{n\frac{2}{2}}\right) $$ Hence $$ k\:=\:\frac{2\pi ^2me^4}{eh^3}….\:\left(xvii\right) $$
$$ E_1\:=-\:\frac{2\pi ^2me^4}{h^2} $$
Putting the value of p, ‘m’: ‘e’ and ‘h’ in the above expression of E1 we get the
Numerical value of E1 as :
$$ E_1\:=-\:\frac{2\:\times \:\left(3.14\right)^2\times \:\left(9.1\times 20^{-28}\right)\:\times \:\left(4.8\times 10^{-10}\right)^4}{\left(6.627\:\times \:\:10^{-2}\right)^2} $$
= -21.79 x 10-12 ergs atom
= -13.6 eV/atom
= – 313.8 Kcal atom
(Since, 1 erg = 6.24 x 1011eV) and 1eV = 23.06Kcals)
Similarly, the values of E2, E3, E4 …. Bto in eV in terms of E1 are calculated by using the following expression:
$$ E_n\:=E_1\times \frac{1}{n^2} $$
$$ E_n\:=-13.6\times \frac{1}{n^2} $$
Limitations of Bohr’s Theory :
2- Bohr’s theory only explained the spectra of H-atom and H-like particles such as He’, etc. but failed the explain the spectra of other atoms having a larger number of electrons.
4- This theory has no explanation for the splitting of spectral lines in an electric field (Stark effect and magnetic effect (Zeeman effect).
5- The magnitude of energies of the stationary states in any atom containing more than one electron can not be accurately calculated by the application of Bohr’s Theory.
6- It does not explain the dual nature of electrons.
Solution According to the equation.
$$ E_n\:=-\frac{2\pi ^2me^4}{n^2h^2} $$
For 4th orbit, n = 4
$$ E_4\:=-\frac{2\pi ^2me^4}{n\frac{2}{4}h^2} $$
$$ =-\frac{2\times \left(3.14\right)^2\times \left(9.109\times 10^{-28}\right)\times \left(4.8\times 10^{-10}\right)^4}{4^2\times \left(6.626\times 10^{-2}\right)} $$
$$ =-1.343\times 10^{-12}+\:rg $$
Or in electron volts
$$ E_A\:=\:-13.6\:\times \:\frac{1}{4^2} $$
$$ =\:-13.6\:\times \:\frac{1}{^{16}} $$
= – 0.85eV
Solution: According to the equation.
$$ r_2=\:\frac{2^2\times \left(6.626\times 10^{27}\right)\:erg^2}{4\times \left(3.14\right)^2\times \left(9.109\times 10^{28}\right)\times 1\times \left(4.8\times 10^{-10}esu\right)^2} $$
= 4 x 0.529 x 10-8cm = g1/2cm3/2 s-1
(erg = g cm2s-2 and e.s.u = g1/2cm3/2 s-1)