1.
(a) This is computed in part (a) of Sample Problem 40-1.
(b) With m
p
= 1.67
× 10
−27
kg, we obtain
E
1
=
h
2
8mL
2
n
2
=
(6.63
× 10
−34
J
·s)
2
8m
p
(100
× 10
12
m)
2
(1)
2
= 3.29
× 10
−21
J = 0.0206 eV .
1.
(a) This is computed in part (a) of Sample Problem 40-1.
(b) With m
p
= 1.67
× 10
−27
kg, we obtain
E
1
=
h
2
8mL
2
n
2
=
(6.63
× 10
−34
J
·s)
2
8m
p
(100
× 10
12
m)
2
(1)
2
= 3.29
× 10
−21
J = 0.0206 eV .