Homework ## Chapter

28: Quantum Physics

Ch 28: 8, 9, 11, 13, 15, 39, 40, 41

28.8 What is the value of Planck's constant in electron - volt - seconds (eV s)?h = 6.626 x 10 ^{ - 34}J sh = 4.14 x 10

_{ }^{- 15}eV s

28.9 What is the wavelength of a 1 eV photon?E = h f f = c

f = c /

E = h c /

= h c / E

= (4.14 x 10

^{ - 15}eV s) (3.0 x 10^{8}m/s) / 1 eV= 1.24 x 10

^{ - 6}m

28.11 Microwaves in household microwave ovens have a wavelength of about 33 cm. What is the energy in one photon of such a microwave?E = h f f = c

f = c /

E = h c /

E = (4.14 x 10

^{ - 15}eV s) (3.0 x 10^{8}m/s) / 0.33 mE = 3.76 x 10

^{ - 6}eV

Remember that we calculated in class that the energy of photons in the visible range was about 2 to 4 eV (also, see problem 28.15 below). Microwaves are of much longer wavelength (0.33 m as compared to about 500 x 10

^{ - 9}m = 5 x 10^{ - 7}m = 0.5 x 10^{ - 6}m or about 10^{6}times as large for the wavelength so about 10^{ - 6}times as small for the energy.

28.13 A gamma ray from a radioactive nucleus has an energy of 21.45 MeV. What is its wavelength?E = h f f = c

f = c /

E = h c /

= h c / E

= (4.14 x 10

^{ - 15}eV s) (3.0 x 10^{8}m/s) / 21.45 MeV= (4.14 x 10

^{ - 15}eV s) (3.0 x 10^{8}m/s) / 21.45 x 10^{6}eV= 5.79 x 10

^{ - 14}m

28.15 What is the energy of the photons in the visible region of the electromagnetic spectrum? (Use l = 550 nm as an average value)?E = h f f = c

f = c /

E = h c /

E = (4.14 x 10

^{ - 15}eV s) (3.0 x 10^{8}m/s) / 550 nmE = (4.14 x 10

^{ - 15}eV s) (3.0 x 10^{8}m/s) / 550 x 10^{ - 9}mE = 2.26 eV

28.39 What is the deBroglie wavelength of a 1.0 - gram paperclip moving at 1.0 m/s? Do you think this length can be detected?= h/p = h / mv = (6.626 x 10

^{ - 34}J s) / [(0.001 kg)(1.0 m/s)]= 6.626 x 10

^{ - 31}mSince the diameter of a nucleus is only about 10

^{ - 14}m, this wavelength will not be measureable at all!

28.40 What is the deBroglie wavelength of a 0.150 - kg baseball when it has been thrown at 28.0 m/s?= h/p = h / mv = (6.626 x 10

^{ - 34}J s) / [(0.150 kg)(28 m/s)]= 1.58 x 10

^{ - 34}mThis wavelength, too, is far too small to be measured or detected.

28.41 What is the wavelength of an electron moving at 1.5 x 10^{7}m/s?= h/p = h / mv = (6.626 x 10

^{ - 34}J s) / [(9.11 x 10^{ - 31}kg)(1.5 x 10^{7}m/s)]= 4.85 x 10

^{ - 11}mThis wavelength is (finally) larger than a nuclear diameter and is a reasonable fraction of typical atomic spacing.

Return to Ch 28, Quantum Mechanics (ToC)(c) Doug Davis, 2002; all rights reserved