7. 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
The energy to be absorbed is therefore
(42 - 12)h2 15(hc)2
"E = E4 - E1 = =
8meL2 8(mec2)L2
15(1240 eV·nm)2
= =90.3 eV.
8(511 × 103 eV)(0.250 nm)2
Wyszukiwarka
Podobne podstrony:
P40p40 4p402p401p406RP 03 Kotowicz p40p409p400p400p40p408więcej podobnych podstron