Page 68 - Electrical Properties of Materials

Exercises 51

(ii) Show that, for high-enough energies, the quantum- 3.8. An electron beam is incident from the x < 0 region
mechanical solution tends to the classical solution. upon the one-dimensional potential shown in Fig. 3.6 with an
energy E =1 eV.
3.4. In electromagnetic theory the conservation of charge is
Without detailed calculations what can you say about the
represented by the continuity equation
reflection coefficient at x =0?
∂N
∇· J =–e ,
∂t
V
where J = current density and N = density of electrons. V = 3 eV
3
Assume that (x, t) is a solution of Schrödinger’s equa-
tion in a one-dimensional problem. Show that, by defining the
x
current density as
i e ∂ ∂ ∗
J(x)=– ∗ – , V = –2 e V
2m ∂x ∂x 2
the continuity equation is satisfied. Fig. 3.6
A one-dimensional potential variation.
3.5. The time-independent Schrödinger equation for the one-
dimensional potential shown in Fig. 3.2 is solved in Sec-
tion 3.6. Using the definition of current density given in 3.9. Solve the time-independent Schrödinger equation for a
the example above, derive expressions for the reflected and two-dimensional rigid potential well having dimensions L x
transmitted currents. Show that the transmitted current is fi- and L y in the x and y directions respectively.
nite when E > V 2 and zero when E < V 2 . Comment on the Determine the energy of the Five lowest lying states when
analogy with Exercise 1.9. L y =(3/2)L x .
3.6. Solve the time-independent Schrödinger equation for the 3.10. The antisymmetric solution of the time-independent
one-dimensional potential shown in Fig. 3.2; Schrödinger equation for the potential well of Fig. 3.4 is given
by eqns (3.32) and (3.34).
V(z)=0, z < 0,
Estimate the energy (a rough estimate will do) of the lowest
V(z)= V 2 , 0 < z < d, antisymmetric state for L =10 –9 mand V 1 = 1.6 × 10 –18 J.
V(z)=0, z > d.
3.11. Solve the time-independent Schrödinger equation for
Assume that an electron beam is incident from the z < 0 the one-dimensional potential well shown in Fig. 3.7, restrict-
region with an energy E. Derive expressions for the reflec- ing the analysis to even functions of x only. The solution
ted and transmitted current. Calculate the transmitted current may be expressed in determinant form. Without expanding the
when V 2 = 2.5 eV, E = 0.5 eV, d =2 Å, and d =20 Å. determinant explain what its roots represent.
3.7. Exercise 3.6 may be solved approximately by assuming
that the wave function in region 2 is of the form ∞ ∞

ψ 2 = Ce –k 2i x V (x)
(where k 2i =Imk 2 ) and then coming to the conclusion (note
that the potential in region 3 is the same as that in region 1 and V 0
therefore k 3 = k 1 )that
E (electron
J 3 –2k 2i d
=e . energy)
J 1
x
How accurate is this approximation? 2b
If the wave function in region 2 is real, as it is assumed in
this exercise, then the quantum mechanical current, as defined 2a
in Exercise 3.4, yields zero.
How is it possible that the equation given above still gives
good approximation for the ratio of the currents whereas the Fig. 3.7
direct use of the proper formula leads to no current at all? A one-dimensional potential well.