Question 39.60: An electron is confined to a narrow evacuated tube of length......

We can use the mc² value for an electron from Table 37-3 (511 × 10³ eV) and hc = 1240 eV · nm by writing Eq. 39-4 as

E_n=\left(\frac{h^2}{8 m L^2}\right) n^2, \quad \text { for } n=1,2,3, \ldots             (39-4)

E_n=\frac{n^2 h^2}{8 m L^2}=\frac{n^2 ( h c )^2}{8( m c^2 ) L^2} .

(a) With L=3.0 \times 10^9 nm, the energy difference is

E_2-E_1=\frac{1240^2}{8 ( 511 \times 10^3 )( 3.0 \times 10^9 )^2} ( 2^2-1^2 ) =1.3 \times 10^{-19} eV \text {. }

(b) Since (n + 1)² – n² = 2n + 1, we have

\Delta E=E_{n+1}-E_n=\frac{h^2}{8 m L^2} ( 2 n+1 ) =\frac{ ( h c ) ^2}{8 (m c^2 ) L^2} ( 2 n+1 ) .

Setting this equal to 1.0 eV, we solve for n:

n=\frac{4\left(m c^2\right) L^2 \Delta E}{(h c)^2}-\frac{1}{2}=\frac{4\left(511 \times 10^3\, eV \right)\left(3.0 \times 10^9\, nm \right)^2(1.0\, eV )}{(1240 \,eV \cdot nm )^2}-\frac{1}{2} \approx 1.2 \times 10^{19} .

(c) At this value of n, the energy is

E_n=\frac{1240^2}{8 ( 511 \times 10^3 )( 3.0 \times 10^9 ) ^2} ( 6 \times 10^{18})^2 \approx 6 \times 10^{18} \,eV .

Thus,

\frac{E_n}{m c^2}=\frac{6 \times 10^{18} eV }{511 \times 10^3 eV }=1.2 \times 10^{13} .

(d) Since E_n / m c^2 \gg 1 , the energy is indeed in the relativistic range.