***CHEMISTRY QUESTION*** ***ELECTRONIC STRUCTURE OF THE ATOM*** ***PLEASE ANSWER
ID: 546801 • Letter: #
Question
***CHEMISTRY QUESTION*** ***ELECTRONIC STRUCTURE OF THE ATOM***
***PLEASE ANSWER CAREFULLY AND COMPLETE ALL PARTS***
The Photoelectric Effect Here are some data collected on a sample of cesium exposed to various energies of light. Electron KE (eV) Electron emitted? Electrons are emitted from the surface of a metal when it's exposed to light. This is called the photoelectric effect. Each metal has a certain threshold frequency of light, below which nothing happens. Right at this threshold frequency, an electron is emitted. Above this frequency, the electron is emitted and the extra energy is transferred to the electron Light energy (eV) 3.87 3.88 3.89 3.90 3.91 no no yes 0.01 yes 0.02 The equation for this phenomenon is Part A where KE is the kinetic energy of the emitted electron, h 6.63 × 10-34 J·s is Planck's constant, v is the frequency of the light, and vo is the threshold frequency of the metal What is the threshold frequency Vo of cesium? Note that 1 eV (electron volt) 1.60 × 10 19 J Express your answer numerically in hertz. Also, since E-hv, the equation can also be written as Hints where E is the energy of the light and E threshold energy of the metal is the Submit My Answers Give UpExplanation / Answer
from given table,
when light energy is 3.89 eV, there is photon emitted but its kinetic energy is 0
So, threshold energy is 3.89 eV
Eo = 3.89 eV
= 3.89*1.60*10^-19 J
= 6.224*10^-19 J
use:
Eo = h*fo
6.224*10^-19 J =(6.626*10^-34 J.s)*fo
fo = 9.39*10^14 Hz
Answer: 9.39*10^14 Hz
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.