3. 24
CHAPTER BASIC IDEAS OF QUANTUM MECHANICS
and see how big it is. Note that h = 1.054,57 1034 J s. Express in units of eV, where one eV
equals 1.602,18 1019 J.
Answer:
E1 =
(1.054,57 1034 J s)2 2
h2 2
=
= 1.506 1018 J
2m2
2 9.109,38 1031 kg (2 1010 m)2
x
or 9.4 eV. The true value is about 4.5 eV. That is in the ball park.
3.5.6.2
Solution pipee-b
Question: Just for fun, plug macroscopic values, m = 1 kg and x = 1 m, into the expression
for the ground state energy and see how big it is. Note that h = 1.054,57 1034 J s.
Answer:
h2 2
(1.054,57 1034 J s)2 2
E1 =
=
= 5.488 1068 J
2
2
2mx
2 1 kg 1 m
or 3.4 1049 eV. That energy is much less than you could ever hope to observe physically. A
single photon of light would it dwarf by 50 orders of magnitude.
3.5.6.3
Solution pipee-c
Question: What is the eigenfunction number, or quantum number, n that produces a macroscopic amount of energy, 1 J, for macroscopic values m = 1 kg and x = 1 m? With that
many energy levels involved, would you see the dierence between successive ones?
Answer: Putting the generic expression for the eigenvalues,
n2 h2 2
En =
2m2
x
equal to 1 J and plugging in the given numbers:
n2 (1.054,57 1034 J s)2 2
= 1J.
2 1 kg 1 m2
Solving for n, you get n = 4.268,64 1033 . Obviously, there is no way to distinguish that many
energy levels. A calculator cannot even display all 34 digits of this number, even if you knew
h to enough digits accuracy to compute 34 digits.