Homework Chapter 02-ER tProblemiD 97807455&offset-next; e Secure httpsisession m
ID: 1086424 • Letter: H
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Homework Chapter 02-ER tProblemiD 97807455&offset-next; e Secure httpsisession masteringchemistryy Chapter 02 ER 4 of 24 equ The Photosictric Effect Here are some data colected on a sample of cesium exposed to various energies of light Light energy Electron emitted? Electron KE Constants (eV) Bectrons are emited from the sueflace of a metal when i's exposed to light. This is called the photlectic efoct Each metal has a certain threshoid trequeney of light, below which nothing happers. Reight at tis threshold frequency, an eledtron is emited. Above this frequeney, the lectron is emtied and the extra energy is (eV) 3.87 3.88 3.89 3.90 3.91 0.01 0.02 The equaton for this phenomenon is KE-ha-ha ehere KE is the kinetic enargy of the Part A constant, vis the frequency of the light, and po is the threshaid frequency of the metal. Aso, since E , the equation Note that 1 eV (electron volt)-1.00 × 10-19 J. Express your answer mumerically in hertz View Avalable Hint(s) can aso be where E is the energy of the light and Eo is the freshoid energy of the metal Ha Part B This quesion will be shown after you complete MacBook Pro G Search or type URLExplanation / Answer
light energy hv = 3.89 eV
kinetic energy KE = 0
hv = hvo
3.89 x 1.60 x 10^-19 = 6.625 x 10^-34 x vo
vo = 9.39 x 10^14 hz
frequency = 9.39 x 10^14 Hz
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