Page 63 - Electrical Properties of Materials

46 The electron

From the continuity of the derivative of the wave function

L L
Ak sin k – Cγ exp –γ = 0. (3.36)
2 2
We have now two linear homogeneous equations in A and C which are
soluble only if the determinant vanishes, that is

L L

cos k
–exp –γ
2 2

= 0, (3.37)

L
L
k sin k –γ exp –γ

2 2
leading to
L

k tan k = γ . (3.38)
2
Thus, k and γ are related by eqn (3.38). Substituting their values from eqns
(3.30) and (3.33) respectively, we get
2m L 2 1/2
E 1/2 tan E =(V 1 – E) 1/2 , (3.39)
2 4
which is a transcendental equation to be solved for E. Nowadays one feeds this
sort of equation into a computer and has the results printed in a few seconds.
But let us be old-fashioned and solve the equation graphically by plotting
the left-hand side and the right-hand side separately. Putting in the numerical
values, we know

m =9.1 × 10 –31 kg, =1.05 × 10 –34 Js,

and we shall take
L –10 –18
=5 × 10 m, V 1 =1.6 × 10 J.
2
If E > V 1 the electron can have As may be seen in Fig. 3.5, the curves intersect each other in three points; so
any energy it likes, but if E < V 1 there are three solutions and that is the lot.
there are only three possible en- To be correct, there are three energy levels for the symmetric solution and a
ergy levels. few more for the antisymmetric solution.
We have at last arrived at the solution of the first quantum-mechanical
problem which deserves literally the name quantum mechanical. Energy is no
longer continuous, it cannot take arbitrary values. Only certain discrete en-
ergy levels are permitted. In the usual jargon of quantum mechanics, it is said:
energy is quantized.
We may generalize further from the above example. The discrete energy
levels obtained are not a coincidence. It is true in general that whenever we try
to confine the electron, the solution consists of a discrete set of wave functions
and energy levels.