25. We can use the mc
2
value for an electron from Table 38-3 (511
× 10
3
eV) and the hc value developed in
problem 3 of Chapter 39 by writing Eq. 40-21 as
E
nx,ny,nz
=
2h
2
8m
n
2
x
L
2
x
+
n
2
y
L
2
y
+
n
2
z
L
2
z
=
(hc)
2
8(mc
2
)
n
2
x
L
2
x
+
n
2
y
L
2
y
+
n
2
z
L
2
z
.
For n
x
= n
y
= n
z
= 1, we obtain
E
1,1
=
(1240 eV
·nm)
2
8(511
× 10
3
eV)
1
(0.800 nm)
2
+
1
(1.600 nm)
2
+
1
(0.400 nm)
2
= 3.1 eV .