How Does the Photoelectric Effect Determine Electron Speed?

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SUMMARY

The discussion focuses on calculating the speed of an electron ejected from a metal surface due to the photoelectric effect when illuminated by light of wavelength 410nm. The work function required to remove the electron is 3.0 x 10^-19 J, and the potential difference between the emitter and collector is 0.5V. The maximum kinetic energy (KEmax) of the electron is calculated as 1.848 x 10^-19 J, leading to a final speed of 419,313.94 m/s using the formula v = √(2KE/m). The calculations and formulas used are confirmed to be correct by other forum participants.

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1. Light of a wavelength 410nm is incident on a metal target and an electron is eject from the surface. The work required to remove this electron from the metal is 3.0 x 10^-19 J. The elctron then travels straight to a collection plate. The different in electric potential between the collector and emitter (target) is .5V (the emitter is at higher potential).

What is the speed of the electron the instant before it strikes the collector

I found that using KEmax = hf - W0, KEmax = 1.848 x 10^-19 J



2. d KE = -|qe Vthreshold |= -|(-1.6x10^-19C)(.5V)|= -8.0 x 10^-20 J



3. v = ROOT (2KE/m) = 419313.94 m/s

i'm not sure if i should be using this formula, nor am I sure how I could apply the kinematics forumlas. Let me know
 
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Welcome to the forums,

Assuming you have done the arithmetic correctly, your answer should be good. It looks to be the right order of magnitude to me :approve:
 
the mathematics are correct, I'm just not sure if I'm using the correct forumlas
 
You're working with the correct formula.
 

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