5. With mp =1.67 × 10-27 kg, we obtain

h2 (6.63 × 10-34 J·s)2
E1 = n2 = (1)2 =3.29 × 10-21 J =0.0206 eV .
8mL2 8mp(100 × 1012 m)2
Alternatively, we can use the mc2 value for a proton from Table 38-3 (938 × 106 eV) and the hc =
1240 eV·nm value developed in problem 3 of Chapter 39 by writing Eq. 40-4 as
n2h2 n2(hc)2
En = = .
8mL2 8(mpc2)L2
This alternative approach is perhaps easier to plug into, but it is recommended that both approaches be
tried to find which is most convenient.