Problems using photon's energy to calculate electron's velocity

  • Thread starter Thread starter mgeoghe2
  • Start date Start date
  • Tags Tags
    Energy Velocity
Click For Summary
SUMMARY

The discussion focuses on calculating the velocity of the fastest electron ejected from a tungsten surface when illuminated by light with a photon energy of 5.8 eV, given a work function of 4.5 eV. The kinetic energy of the ejected electron is determined using the equation KEmax = hf - work function, leading to KEmax = 5.8 eV - 4.5 eV = 1.3 eV. To find the velocity, the mass of the electron (9.109e-31 kg) is used in conjunction with the equation v^2/c^2 = 2 * KE/mc^2, where c is the speed of light.

PREREQUISITES
  • Understanding of the photoelectric effect and work function
  • Knowledge of kinetic energy equations
  • Familiarity with electron mass and energy units (eV, kg, MeV)
  • Basic grasp of relativistic equations
NEXT STEPS
  • Study the photoelectric effect in detail, focusing on the role of work function
  • Learn how to convert between energy units (eV, MeV, Joules)
  • Explore relativistic effects on electron velocity calculations
  • Practice problems involving kinetic energy and electron dynamics
USEFUL FOR

Physics students, educators, and anyone interested in quantum mechanics and the photoelectric effect will benefit from this discussion.

mgeoghe2
Messages
2
Reaction score
0
ive got this problem in which i am given:
the work function of tungsten = 4.5 eV
light with photon energy 5.8 eV illuminates surface

and i need to figure out what the velocity of the fastest electron ejected from the surface is.

so i see work function and 'fastest' and i think, ok KEmax = hf - work function. and KE is just .5mv^2. the problems i run into are these:
where do i get f? or hf?
i see that E = hf , so i think, ok i will use 5.8 -4.5 = KEmax. but then i run into another problem.
which mass do i use? i got really screwed up with the units... i can use 9.109e-31 kg but i don't know if and eV is kg m^2/s^2. i am also given the mass of an electron as .511 MeV/c^2 but i don't really understand what that means. is it .511e6 or .511e6/(3e8)^2??
if you have any insightful comments, i would be very grateful.
 
Physics news on Phys.org
Since you know the rest energy of electron you can find the speed from

[tex]\frac {v^2}{c^2} = \frac {2 \times 1.3 eV}{m c^2}[/tex]
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
10
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
13K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
30K
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K