chap38 - 38.1: Hz 10 5.77 m 10 5.20 s m 10 3.00 λ 14 7 8...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 38.1: Hz 10 5.77 m 10 5.20 s m 10 3.00 λ 14 7 8 × = × × = =- c f eV. 2.40 J 10 3.84 ) s m 10 (3.00 s) m kg 10 (1.28 s m kg 10 1.28 m 10 5.20 s J 10 6.63 λ 19 8 27 27 7 34 = × = × ⋅ × = = ⋅ × = × ⋅ × = =----- pc E h p 38.2: a) eV. 10 7.49 J 0.0120 s) 10 (20.0 W) (0.600 16 3 × = = × =- Pt b) eV. 1.90 J 10 3.05 λ 19 = × = =- hc hf c) . 10 94 . 3 16 × = hf Pt 38.3: a) Hz. 10 5.91 s J 10 6.63 eV) J 10 (1.60 eV) 10 (2.45 20 34 19 6 × = ⋅ × × × = = ⇒ =-- h E f hf E b) m. 10 5.08 Hz 10 5.91 s m 10 3.00 λ 13 20 8- × = × × = = f c c) λ is of the same magnitude as a nuclear radius. 38.4: hc W hc P hf P dN dE dt dE dt dN m) 10 (2.48 ) (12.0 λ ) ( ) ( 7- × = = = = sec. photons 10 1.50 19 × = 38.5: φ- = hf mv 2 max 2 1 J 10 3.04 eV J 10 (1.60 eV) (5.1 m 10 2.35 s J 10 (6.63 20 19 7 34 ) )---- × × × ⋅ × =- = c . s m 10 2.58 kg 10 9.11 J) 10 2(3.04 5 31 20 max × = × × = ⇒-- v 38.6: ×- × =- =- =- m 10 2.72 Hz 10 1.45 λ 7 15 c h hc hf hf E φ eV. 1.44 J 10 2.30 19 = × =- 38.7: a) Hz 10 5.0 λ 14 × = = c f b) Each photon has energy J. 10 .31 3 19- × = = hf E Source emits s photons 10 2.3 photons/s) 10 (3.31 ) s J (75 so s J 75 20 19 × = ×- c) No, they are different. The frequency depends on the energy of each photon and the number of photons per second depends on the power output of the source. 38.8: For red light nm 700 λ = eV 1.77 J 10 1.6 1eV J 10 2.84 m) 10 (700 ) s m 10 (3.00 s) J 10 (6.626 λ 19 19 9 8 34 = × × = × × ⋅ × = = =---- hc hf φ 38.9: a) For a particle with mass, . 4 means 2 . 2 1 2 1 2 2 K K p p m p K = = = b) For a photon, . 2 means 2 . 1 2 1 2 E E p p pc E = = = 38.10: φ- = hf K max Use the information given for : find to nm 400 λ φ = J 10 3.204 eV) J 10 (1.602 eV) (1.10 m 10 400 ) s m 10 (2.998 s) J 10 (6.626 19 19 9 8 34 max---- × = ×- × × ⋅ × =- = K hf φ Now calculate : nm 300 λ for max = K eV 2.13 J 10 3.418 J 10 3.204 m 10 300 ) s m 10 (2.998 s) J 10 (6.626 19 19 9 8 34 max = × = ×- × × ⋅ × =- =---- φ hf K 38.11: a) The work function eV λ eV- =- = hc hf φ J. 10 7.53 V) (0.181 C) 10 (1.60 m 10 2.54 s) m 10 (3.00 s) J 10 (6.63 19 19 7 8 34---- × = ×- × × ⋅ × = ⇒ φ The threshold frequency implies φ φ hc hc hf th th = ⇒ ⇒ = th λ λ m. 10 2.64 J 10 7.53 ) s m 10 (3.00 s) J 10 (6.63 λ 7 19 8 34 th--- × = × × ⋅ × = ⇒ b) eV, 4.70 J 10 7.53 19 = × =- φ as found in part (a), and this is the value from Table 38.1. 38.12: a) From Eq. (38.4), V. 2.7 V 2.3 m) 10 (2.50 s) m 10 (3.00 s) eV 10 (4.136 λ 1 7 8 15 =- × × ⋅ × = - =-- φ hc e V b) The stopping potential, multiplied by the electron charge, is the maximum kinetic energy, 2.7 eV....
View Full Document

This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

Page1 / 24

chap38 - 38.1: Hz 10 5.77 m 10 5.20 s m 10 3.00 λ 14 7 8...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online