Here are some data collected on a sample of cesium exposed to various energies o

Question

Here are some data collected on a sample of cesium exposed to various energies of light...

Light energy (eV) -- Electron emitted? -- Electron KE (eV)

3.87                                       no                                 —

3.88                                       no                                 —

3.89                                      yes                                 0

3.90                                      yes                                 0.01

3.91                                      yes                                 0.02

Part C

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.9×10^15Hz ?

Express your answer in joules to three significant figures... Hint* Its close too (6.5*10^-19 J) but I keep on getting this one wrong...

Explanation / Answer

f = 1.9*10^15 Hz
energy of this photon,
E = h*f
= (6.626*10^-34 * 1.9*10^15)
= 1.259*10^-18 J

Eo = 3.89 eV
= 3.89 * 1.602*10^-19 J
= 6.232*10^-19 J

use
KE = E - Eo
= (1.259*10^-18 J) - (6.232*10^-19 J)
= 6.358*10^-19 J

Answer: 6.4*10^-19 J

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