13. We can use the mc2 value for an electron from Table 38-3 (511 × 103 eV) and the hc value developed in
problem 3 of Chapter 39 by writing Eq. 40-4 as
n2h2 n2(hc)2
En = = .
8mL2 8(mc2)L2
(a) With L =3.0 × 109 nm, the energy difference is
12402
E2 - E1 = 22 - 12 =1.3 × 10-19 eV .
8(511 × 103)(3.0 × 109)2
(b) Since (n +1)2 - n2 =2n +1, we have
h2 (hc)2
"E = En+1 - En = (2n +1) = (2n +1) .
8mL2 8(mc2)L2
Setting this equal to 1.0 eV, we solve for n:
4(mc2)L2"E 1
n = -
(hc)2 2
4(511 × 103 eV)(3.0 × 109 nm)2(1.0eV) 1
= -
(1240 eV·nm)2 2
H" 12 × 1018 .
(c) At this value of n, the energy is
2
12402
En = 6 × 1018 H" 6 × 1018 eV .
8(511 × 103)(3.0 × 109)2
(d) Since
En 6 × 1018 eV
= 1 ,
mc2 511 × 103 eV
the energy is indeed in the relativistic range.
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