Page 435 - Schaum's Outline of Theory and Problems of Applied Physics

420 ATOMIC PHYSICS [CHAP. 34

The kinetic energy of the electrons is
4
KE = (1.5 × 10 eV)(1.6 × 10 −19 J/eV) = 2.4 × 10 −15 J
1
2
Since KE = mv , the velocity of the electrons is
2

2(KE) (2)(2.4 × 10 −15 J)
7
v = = = 7.26 × 10 m/s
m 9.1 × 10 −31 kg
The electron wavelength is therefore
h 6.63 × 10 −34 J·s −11
λ = = = 1.0 × 10 m
mv (9.1 × 10 −31 kg)(7.26 × 10 m/s)
7
UNCERTAINTY PRINCIPLE
A consequence of the wave nature of moving bodies is the uncertainty principle: It is impossible to determine
both the exact position and the exact momentum of a body at the same time. If x is the uncertainty in position
and (mv) is the uncertainty in momentum, then
h
x (mv) ≥
2π
where the symbol ≥ means “equal to or greater than.” Since Planck’s constant h is so small, the uncertainty
principle is significant only for very small bodies such as elementary particles.

SOLVED PROBLEM 34.3
The position of a certain electron is determined at a certain time with an uncertainty of 10 −9 m. (a) Find
the uncertainty in the electron’s momentum. (b) Find the uncertainty in the electron’s velocity and, from
this, the uncertainty in its position 1 s after the original measurement was made.
h 6.63 × 10 −34 J·s
(a) (mv) = = = 1.06 × 10 −25 kg·m/s
2π x (2π)(10 −9 m)

(mv) 1.06 × 10 −25 kg·m/s
5
(b) v = = = 1.2 × 10 m/s
m 9.1 × 10 −31 kg
Hence the uncertainty in the electron’s position 1 s later is
5
5
x = t v = (1s)(1.2 × 10 m/s) = 1.2 × 10 m

which is 120 km!

SOLVED PROBLEM 34.4
Most atoms are a little over 10 −10 m in radius. (a) Find the uncertainty in the momentum of an electron
whose position is known to within 10 −10 m. (b) Find the corresponding uncertainty in the electron’s
kinetic energy. What is the significance of this figure?
(a) The momentum uncertainty is

h 6.63 × 10 −34 J·s −24
(mv) = = = 1.06 × 10 kg·m/s
2π x (2π)(10 −10 m)
2
1
2
(b) Since KE = mv = 1/(mv) /2m,
2
1 (1.06 × 10 −24 kg·m/s) 2
2
KE = ( mv) = = 6.1 × 10 −19 J = 3.8eV
2m (2)(9.11 × 10 −31 kg)