{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ProbSet2Solutions

# ProbSet2Solutions - Solutions to Problem Set 2 1.2 1.4...

This preview shows pages 1–2. Sign up to view the full content.

Solutions to Problem Set 2 1.2 radio waves < IR rays < visible light < UV rays 1.4 (a) 2.99792 x 10 8 m.s -1 = ( λ )(7.2 x 10 14 s -1 ) λ = (2.99792 x 10 8 m.s -1 )/(7.1 x 10 14 s -1 ) = 4.2 x 10 -7 m = 420 nm (b) 2.9972 x 10 8 m.s -1 ) = ( λ )(2.0 x 10 18 s -1 ) λ = (2.99792 x 10 8 m.s -1 )/(2.0 x 10 18 s -1 ) = 1.5 x 10 -10 m = 150 pm 1.12 (a) false. UV photons have higher energy than infrared photons. (b) false. The kinetic energy of the electron is directly proportional to the energy ( and hence frequency) of the radiation in excess of the amount of energy required to eject the electron from the metal surface. (c) true. 1.14 E work function = (4.37 eV)(1.6022 x 10 -19 J.eV -1 ) = 7.00 x 10 -19 J E kinetic = ½ mv2 = 0.5 (9.10939 x 10 -31 kg) ( 1.5 x 10 6 m.s -1 ) 2 = 1.02 x 10 -18 J E total = E work function + E kinetic = 1.72 x 10 -18 J E = h ν and c = νλ λ = hc/E = (6.626 x 10 -34 Js)(3.00 x 10 8 m.s -1 )/(1.72 x 10 -18 J) = 1.16 x 10 -7 m = 116 nm 1.18 mass of one He atom = Molar mass of He / Avogadro’s number = (4.00 g.mol -1 )/(6.022 x 10 23 atoms. mol -1 ) = 6.64 x 10 -24 g = 6.64 x 10 -27 kg From the de Broglie relationship, λ = h/mv = (6.626 x 10 -34 Js)/(6.64 x 10 -27 kg)(1230 m.s -1 ) = 8.11 x 10 -11 m 1.22 The observed line is the third lowest energy line. The frequency of the given line is

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

ProbSet2Solutions - Solutions to Problem Set 2 1.2 1.4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online