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Unformatted text preview: Physics 208, Assignment 11, Solutions, 4/15/09, RT Problem 1. eV stop [eV] hf [eV] hf V + _ T C phototube hf 2 1 3 0.5 1.0 1.5-0.5-1.0-1.5 (g)- (a,b) After steady illumination the side of the voltmeter labelled - will be just negative enough (due to previously emitted electrons) to repel further electrons. (c) Using given data and hf [eV] = 1240 / [nm], a table can be filled in: The data is consistent with color hf V stop hf- eV stop nm eV V eV violet 436 2.84 1.34 1.50 blue 488 2.54 1.04 1.50 yel/gr 546 2.27 0.77 1.50 orange 610 2.03 0.53 1.50 hf- eV stop = 1 . 50 V . (1) which defines the work function . (d) Since emitted electrons cannot have negative kinetic energy, neg- ative values of eV stop are not allowed in Eq. (1). The intercept with the horizontal axis determines the cut-off frequency to be f = 1 . 50 / 4 . 14 10- 15 eV.s = 3.62e14Hz. This is near infrared. (e) The corresponding wavelength is = 1240 / 1 . 50 = 826 nm. This is the longest wavelength capable of ejecting electrons. All shorter wavelengths ( < ) which includes visible, UV, and beyond, can eject electrons. 1 (f) Changing the light intensity would not change the stopping voltage, but it would affect the time required after voltage is applied for the collector to charge up....
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This note was uploaded on 09/01/2009 for the course PHYS 208 taught by Professor Amadeuri during the Spring '08 term at Cornell University (Engineering School).
- Spring '08