Finding the kinetic energy of an electron

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SUMMARY

The discussion centers on calculating the kinetic energy of an electron emitted from a magnesium target when exposed to light. The relevant equation for kinetic energy (KE) is established as KE = pf - phi, where phi represents the work function of magnesium at 3.7 eV. The user correctly identifies the need to calculate the energy from the electron's velocity of 2 * 10^6 m/s using the formula E = 0.5 * m * v^2, confirming the approach to find the necessary wavelength of light for photoelectron emission.

PREREQUISITES
  • Understanding of kinetic energy calculations in physics
  • Familiarity with the photoelectric effect
  • Knowledge of the work function concept
  • Basic proficiency in using Planck's constant (h = 6.63 * 10^-34 J·s)
NEXT STEPS
  • Research the photoelectric effect and its applications in modern physics
  • Learn how to calculate the wavelength of light using the energy of emitted electrons
  • Explore the relationship between velocity and kinetic energy in particle physics
  • Study the implications of work function variations in different materials
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism and quantum mechanics, as well as educators seeking to explain the photoelectric effect and kinetic energy calculations.

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Homework Statement



This is a basic photoelectric type question.

What wavelength of light is necessary to produce photo-electrons of speed 2 * 10^6 m/s with a magnesium target?

Homework Equations



KE = pf - phi

phi = 3.7 eV
h = 6.63 * 10 ^-34

The Attempt at a Solution



I don't know how to find the energy of an electron from the velocity. Once I have that I think I can figure it out
 
Last edited:
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I think I figured it out! E = .5 * m * v^2
 
Your equations are right =]
 

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