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 .