This
preview
has intentionally blurred sections.
Sign up to view the full version.
This
preview
has intentionally blurred sections.
Sign up to view the full version.
This
preview
has intentionally blurred sections.
Sign up to view the full version.
Unformatted text preview: n = 1 and n = 2 we get: E 1 =-k 2 Z 2 e 4 m π 2 ~ 2 =-(8 . 99 × 10 9 Jm/C 2 ) 2 (26 2 )(1 . 60 × 10-19 C ) 4 (273)(9 . 11 × 10-31 kg ) 2(1 . 05 × 10-34 J · s ) 2 =-4 . 04 × 10-13 J =-2 . 52 × 10 6 eV E 2 =-k 2 Z 2 e 4 m π 8 ~ 2 =-(8 . 99 × 10 9 Jm/C 2 ) 2 (26 2 )(1 . 60 × 10-19 C ) 4 (273)(9 . 11 × 10-31 kg ) 8(1 . 05 × 10-34 J · s ) 2 =-1 . 01 × 10 6 J =-6 . 31 × 10 5 eV Part c) Finally, the energy of the photon emitted upon transition from level 2 to level 1 is the difference between the energies of the two levels: E = E 2-E 1 = 1 . 89 × 10 6 eV 1...
View
Full Document
- Spring '08
- LIN
- Energy, Kinetic Energy, Potential Energy, Quantum Physics, 2 K, 31 kg, 13 J
-
Click to edit the document details