Consider an electron of charge ‘e’ revolving around a nucleus of charge ‘ze’, where ‘z’ is the atomic number and ‘e’ the charge on a proton. Let ‘m’ be the mass of the electron, ‘r’ be the radius of the orbit and ‘v’ the tangential velocity of the revolving electron. The electrostatic force of attraction between the nucleus and the electron (coulomb’s law) = ze × e / r2 .
The centrifugal force acting on the electron = mv2 / r
Bohr assumed that these two opposing forces must be balancing each other exactly to keep the electron in orbit. Thus,
Z2/ r2 = mv2/r
For hydrogen z = 1, therefore,
e2/ r2 = mv2/r -------(1)
Multiplying both sides by ‘r’
e2/r = mv2 -------(2)
According to one of the postulates of the Bohr’s theory, angular momentum of the revolving electron is integral multiple of h/2π.
mvr = nh/2π
v = nh/2πmr
Substitute the value of ‘v’ in equation (2),
e2/r = m﴾ nh/2πmr)2
So, r = n2h2/4π2me2 -------(3)
Well written information.Thanks
ReplyDelete