# APPLICATION OF BOHR’S THEORY

by • 01/08/2013 • GeneralComments (0)433

APPLICATION OF BOHR’S THEORY

Consideration:

Consider an atom of hydrogen with one election and one proton.  Proton is present inside the nucleus with a charge e+ while the charge of electron is e+. The distance between electron and 6 protons is “r” which is called atomic radius, as shown in figure.

DERIVATION:

Since electron is moving in a circular path so it posses centripetal force.

Fe = Z mV² / r

This centripetal force is provided to electron by the attraction between electron and proton that is

Fe = Z q1q2 / r²

Fe = Z (e)(e)/ r²

Fe = Z e²/ r²

Since,

Fc = Fe

mV² / r = Z e²/ r²

mV² = Z e²/ r —–(1)

According to Bohr’s theory

mvr = nh/2π

V = nh/ 2πmr

Put the value of v in eq (1)

mV² = Z e²/ r

m (nh/2π) ² = Z e²/ r

mn²h² / 4 π ²m²r² = Z e²/ r

n²h² / 4 π ²mr = Z e²

r = n²h² / 4 π ²m Z e²

For

n = 1

h = 6.625 x 10-27 eng. Sec

π = 3.14

m = 9.11 x 10-28 gm

e = 4.802 x 10-10 esu

Z = 1

R1= 0.529˚ A or 0.529 x 10-8 cm

So,

rn = n²r¹

ENERGY OF ORBITS:

In any orbit the total energy of electrons are equal to the sum of kinetic energy and potential energy due to position up to radius “I” i.e.

T. E = K. E + P. E

E = ½ mv² + Ze²/r

E = ½ mv² – Ze²/r — (1)

According to Bohr’s theory

rnV² = Ze²/r

So equation 1

E = ½ mv² – Ze²/r

E = -½ Z e²/r

By putting the value of r which is

r = n²h²/ 4 π²mZe²

E = -1/2 Ze / n²h²/ 4 π²mZe²

E = -1/2 4Z²e²m π ² / n²h²

E = -2m π ²Z²e² / n²h²

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