CHEM 20A. Winter 2001
Assignment 4. SOLUTIONS.
9.1.a) If the wave length is much longer than the object, it is like having the wave length
(λ) much longer than the separation between slits (d). For the 2-slit experiment,
nλ = dsinθ,
where θ is the angle at which the beam is measured (scattering angle).
The first diffraction maximum comes at n = 1, so if d/λ < 1, it is impossible to satisfy the
interference condition. Why? On the other hand if λ is too small, d/λ >> 1, then the first
interference maximum comes at such a small angle that it cannot be detected (it is
unresolved). Why?
b) E = (1/2)mv2 = (1/2)(mv)2/m = (1/2m)(h/λ)2
= (1/2)(1/9x10-31kg)(6.6x10-34J s/10-10m)2(1/1.6x10-19 J eV-1)
= 1.4x102 eV
c) Same as above but mass is 1800 times larger. Since the energy above varied as m-1,
the energy of the neutron is 1800 times less than than for the electrons.
E = (1.4x102/1.8x103)eV = 0.08 eV.
9.2. This is done in many texts. You may come see DK about it if your are interested.
9.3. a). This is done in many texts. You may come see DK about it if you are interested.
b) .This is done in many texts. You may come see DK about it if you are interested.
c) If d is the distance between planes, and a bright spot is observed at an angle θ,
then the distance between planes can be determined by the formula, d = (λl/2sinθ).
d) See problem 9.1.
10.2. We ask first what constants can possibly enter the problem? Since it is a quantum
mechanical problem it must contain Planck's constant, h. The kinetic energy must
depend upon the electron mass, m. The potential energy must depend upon (Ze2/4πεo).
Why? There are no other constant that can enter this problem. Why?
If there is a length it must depend upon the constants above. Why? Let us call the length,
az. Thus
az = a function of the three constants above.
Furthermore, the dimensions on the left of this equation are length (L) and so the
dimensions on the right must be length (L). So
aZ is proportional to (m)b(h)c(Ze2/4πεo)d
where b,c, and d are dimensionless numbers that we wish to determine.
The dimensions of mass are clearly mass {M}}.
The dimensions of h can readily be determined since hν = energy = {ML2T-2} where {T}
is time. Since the dimensions of ν are {T-1}, it follows that the dimensions of h are
{ML2T-1}.
The dimensions of (Ze2/4πεo) can readily determined since (Ze2/4πεo) divided by a length
is an energy. (Coulomb's law). Thus (Ze2/4πεo) is an energy times a length, i.e.,
{ML2T-2L} = {ML3T-2}
Rewriting the equation just in terms of dimensions we have
{L} = {M}b{ML2T-1}c{ML3T-2}d
