Solutions Homework 5 Chapter 3

Solutions Homework 5 Chapter 3 - Solutions Homework 5...

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

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions Homework 5 Chapter 3 3-31. KM “MW {4 13g ' . i _ “Mai—i}— m w; x m“ .32. ” 4,13%; 21343 ex: 1 Fzr v 1 in“ PM) 2 “- M— II: is” 31:: s w" m if; M 1. ' ‘ A E z A 1 [1240 9&7 r 32:.” W 1: M :55 fi {Ll-3% 4&3 mm { ] ‘J he 1240 eV' - nm 2 A: Z W If 2 2.0 e\-" = hf — (3) K + (p 2.0 elm-T + 4.59 eV" = h = 4.136 x 10-15 eV-s (:1 = 4.5-9 eV f = 1.59 x 1015 Hz 39. eVm = 11 6101.1 — (3? and El’hg = lief-Q2 — to. Subtracting these equations and rearranging we find e (1432 — V01) e(2.3 v — 1.0 V") _ _ 4.40 x 10—15 eV .5 4% — (3-3” X 103 “155) (m — m} 11- This is about 60/:- from the accepted value. For the work function we use the first set of data If the second set should give the same result): he _, {4.40 x 10—1'5 eV. 5} {3.00 x 11]8 mfs) (.3 — Elm — A1 260 x 10—9 n1 1.0 eV — 4.1 eV 48. Use the Compton scattering fornmla but with the protons mass and 9 = 00°: h )2 F16 1240 eV - nm A). _ 1 . El _ _ _ _ 1.32 f me ( COS ) me 31102 938.27 MeV m 52. I he A —/\ l /\le1 c059)— E lAcfl cost?) _ 1240 e\-" -nrn _ 700 x 103 e\-" be 1240 eV - 11m = _ = — = 2.45 10-51:“ = 24’ k V )1’ 5.03 x 10—3 nm X e e B}; conservation of energy A’ + {2.43 x 10—3 mm) (1 — cosllfl”) = 5.03 pm Elf Kg = E—E’ = 700 lieV —246 lieV = 45-4 lieV (agrees with If formula in the previous problem] hf 6? 700 keV 1100 1 -c — I 1 — '.' 8' F 1 meg] t‘1n(2) [1 Sllkexlr]t111( 2 ) 3.3 ~11 (:1 = 16.50 From Problem 51 cot (in — ’V' Fair A 12%| 91’ » mm a? m 71,64 :93? 1f the ‘A'amlmlfi'ih inked fit; A x 13333 um} A: :2 m Iilil far 324%} 935 - 1:111; , _ 5-" z .4 m y = w————-.. ‘ w 41:“; a: m 43:; A; ‘ A ‘j 133A51’m.) " g F ha. >x 490 Light of wavelength 400 um liberates photoelectrons {Tom a certain metal. The photoelectrons now enter a uniform magnetic field of 104 T. The electrons move normal to the field lines so that they travel circular paths. The largest circular path has a radius of 5.14 cm. Find the work fimction for the metal. PM a t— .T— ,IU“ :- W N“ ha ,_ l U“:- W '7‘" " —- 7”” n L 6b we“ -- ME?) >\ 2. q '2; 4? :‘I W- 5.7!,(sa T Hooxw’ -t‘l "I? .b 3.7! X10 .7" .7: Ll-fiééKloj '- Lifklorfifzfi~79 elf ‘— ...
View Full Document

This note was uploaded on 03/30/2008 for the course PHYS 222 taught by Professor Adair during the Spring '08 term at Texas A&M.

Page1 / 4

Solutions Homework 5 Chapter 3 - Solutions Homework 5...

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