Radius and Energy Levels of Hydrogen Atom:

Consider an electron of mass ‘m’ and charge ‘e’ revolving around a nucleus of charge Ze (where, Z = atomic number and e is the charge of the proton) with a tangential velocity v. r is the radius of the orbit in which electron is revolving.

By Coulomb’s Law, the electrostatic force of attraction between the moving electron and nucleus is Coulombic force = KZe2/r2

K = 1/4π∈o  (where ∈o is permittivity of free space)

K = 9 x109 Nm2 C–2

In C.G.S. Units, value of K = 1 dyne cm2 (esu)–2

The centrifugal force acting on the electron is mv2/r

Since the electrostatic force balances the centrifugal force, for the stable electron orbit.

mv2/r= KZe2/r2                                         … (1)

or     v2 =   KZe2/mr                                          … (2)

According to Bohr’s postulate of angular momentum quantization, we have

mvr = nh/2π

v = nh/2πmr

v2 = n2h2/4π2m2r2                                         … (3)

Equating (2) and (3) 

KZe2/mr = n2h2/4π2m2r2

Solving for r we getr = n2h2/4π2mKZe2

Where n = 1, 2, 3 - - - - - ∞ 

Hence only certain orbits whose radii are given by the above equation are available for the electron. The greater the value of n, i.e., farther the energy level from the nucleus the greater is the radius.

The radius of the smallest orbit (n=1) for hydrogen atom (Z=1) is ro.

ro = n2h2/4π2me2K = 12 x (6.626 x 10-34)2 / 4x(3.14)2 x9x10-31 x (1.6x10-19)2x9x109 =5.29 x 10–11 m=0.529 Å

Radius of nth orbit for an atom with atomic number Z is simply written as

rn = 0.529 x n2/z Å

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