{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 9&10 solutions (Ch 11,28)

# HW 9&10 solutions (Ch 11,28) - Physics 6C HW#9 and#10...

This preview shows pages 1–3. Sign up to view the full content.

Physics 6C HW#9 and #10 solution June 5, 2008 11.34 a) The energy of the photon is found as E = E f - E i = - 13 . 606 eV n 2 f - - 13 . 606 eV n 2 i here n f = 2 and n i = 3 = 13 . 606 eV ( 1 4 - 1 9 ) = 1.89 eV b) E = hc λ λ = 6 . 626 × 10 - 34 J · s (2 . 998 × 10 8 m/s ) 1 . 89 eV (1 . 602 × 10 - 19 J/eV ) = 656 nm c) f = c λ = 3 × 10 8 m/s 6 . 56 × 10 - 7 m = 4 . 57 × 10 14 Hz 11.37 Δ E = ( - 13 . 6 eV )( 1 n 2 f - 1 n 2 i ) Where for Δ E > 0 we have absorption and for Δ E < 0 we have emission. 1. for n i = 2 and n f = 5, Δ E = 2 . 86 eV(absorption) 2. for n i = 5 and n f = 3, Δ E = - 0 . 967 eV(emission) 3. for n i = 7 and n f = 4, Δ E = - 0 . 572 eV(emission) 4. for n i = 4 and n f = 7, Δ E = 0 . 572 eV(absorption) a) E = hc λ so the shortest wavelength is emitted in tran- sition 2 . b) The atom gains most energy in transition 1 . c) The atom loses energy in transition 2 and 3 . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
11.39 a) r n = a 0 n 2 where a 0 = 0 . 0529 nm r 2 = 4 a 0 = 0.212 nm b) p = m e v 2 = s m e k e e 2 r 2 = 9 . 95 × 10 - 25 Kg m/s c) l 2 = | r 2 × p | = m e v 2 r 2 = 2 . 11 × 10 - 34 kg · m 2 /s d) K 2 = p 2 2 2 m e = 5 . 43 × 10 - 19 J = 3.40 eV e) U 2 = - k e e 2 r 2 = - 1 . 09 × 10 - 18 J × 1 eV 1 . 60 × 10 - 19 = -6.80 eV f) E 2 = K 2 + U 2 = 3 . 40 eV - 6 . 80 eV = -3.40 eV 28.21 a) For an electron, λ = h p K = p 2 2 m e p = p 2 m e K
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

HW 9&10 solutions (Ch 11,28) - Physics 6C HW#9 and#10...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online