Page 436 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 34] ATOMIC PHYSICS 421

In order for this to be the uncertainty in electron’s KE, the KE itself must be greater, which in fact is true for
all atomic electrons.

BOHR MODELOF THE HYDROGEN ATOM
The hydrogen atom consists of a single electron and a single proton, which is the nucleus. In the classical model,
the electron is imagined to circle the proton in an orbit such that the electric attraction of the proton provides the
required centripetal force. The flaw in this model is that, according to electromagnetic theory, because the electron
is accelerated, it must radiate electromagnetic waves and thus will lose energy until it spirals into the proton.
In the Bohr model, it is postulated that stable orbits exist in which the angular momentum of the electron
is a multiple of h/(2π), that is, that the angular momentum is nh/(2π), where n = 1, 2, 3,.... This postulate
is equivalent to requiring that each orbit be a whole number of de Broglie wavelengths in circumference. If n is
the quantum number of an orbit, then the orbit radius is
2
r n = n r 1 n = 1, 2, 3,...
where r 1 , the radius of the smallest orbit, is 5.3 × 10 −11 m.

ENERGY LEVELS
The total energy (kinetic energy plus electric potential energy) of a hydrogen atom whose electron is in the nth
orbit is
E 1
E n = n = 1, 2, 3,...
n 2
where E 1 =−13.6eV =−2.18 × 10 −18 J. The permitted energies of an atom are called its energy levels.
The energy levels are all negative, which means that the electron does not have enough energy to escape from
the proton. The lowest level, corresponding to n = 1, is the ground state; higher levels are excited states.As n
increases, E n approaches zero; when E n = 0, the electron is no longer bound to the proton and the atom breaks
up. The work needed to remove an electron from an atom in its ground state is called the ionization energy;itis
13.6 eV in the case of hydrogen.

SOLVED PROBLEM 34.5

Find the energies of the first three excited states of the hydrogen atom.
2
Since E n = E 1 /n and E 1 =−13.6 eV, we have
E 1 −13.6eV
E 2 = = =−3.4eV
2 2 4
E 1 −13.6eV
E 3 = = =−1.51 eV
3 2 9
−13.6eV
E 1
E 4 = = =−0.85 eV
4 2 16
SOLVED PROBLEM 34.6
Find the velocity of the electron in a ground-state hydrogen atom according to the Bohr model.

In the Bohr model, the de Broglie wavelength λ = h/(mv) of the electron in the n = 1 state is equal to the orbit
circumference of 2πr 1 . Hence
h
λ = = 2πr 1
mv
h 6.63 × 10 −34 J·s
6
v = = −31 −11 = 2.2 × 10 m/s
2πmr 1 (2π)(9.1 × 10 kg)(5.3 × 10 m)