How Do You Calculate Photoelectron Emission from Sodium Surfaces?

Click For Summary
SUMMARY

The calculation of photoelectron emission from sodium surfaces involves determining the energy of ejected electrons when illuminated by light of a specific wavelength. For sodium, the threshold wavelength is 683 nm, corresponding to a threshold frequency of 6E14 Hz and an electron binding energy of 2.91E-19 J. When illuminated by 500 nm light, the kinetic energy of the ejected electrons is calculated to be 1.1E-19 J. Given an intensity of 2.0 W/m² and a surface area of 2E-4 m², the estimated number of emitted photoelectrons is 3.6E13, although a discrepancy with a reference book suggests a correct value of 1.0E13, indicating a calculation error in using the kinetic energy instead of the energy of incident photons.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Knowledge of photon energy calculations using wavelength
  • Familiarity with basic principles of light intensity and surface area
  • Ability to perform energy and frequency conversions
NEXT STEPS
  • Learn about photon energy calculations using the formula E = hc/λ
  • Study the principles of the photoelectric effect in detail
  • Explore the relationship between light intensity and photoelectron emission
  • Investigate common errors in photoelectron calculations and their corrections
USEFUL FOR

Students studying quantum mechanics, physicists interested in the photoelectric effect, and educators teaching concepts related to light and electron interactions.

kasse
Messages
383
Reaction score
1

Homework Statement



The wavelength threshold for photoelectric emission from a sodium surface is 683 nm. Calculate the energy of the electrons which are ejected when a sodium surface is illuminated by light of wavelength 500 nm. If the intensity of the light is 2.0 W/m^2 and if 1 per cent of incident photons produce photoelectrons, estimate the number of electrons emitted per second from a sodium surface of area 2*10^-4 m^2 when it is illuminated by light of the wavelength 500 nm.

2. The attempt at a solution

The threshold wavelength corresponds to a threshold frequency of 6E14 Hz, which in turn corresponds to a electron binding energy of 2.91E-19 J.

This means that the energy of the electrons which are ejected is K = 1.1E-19J.

Next, we observe that since the surface area is 2E-4 m^2, the total effect is 4E-4 W, and hence 4E-4 J of energy hit the surface each second.

Since 1 per cent of this energy produces photoelectrons, the total photoelectron energy per second is 4E-6 J.

Since each photoelectron has energy 1.1E-19, my estimated number of photoelectrons is 4E-6/1.1E-19 = 3.6E13 photoelectrons.

My book says 1.0E13. What have I done wrong?
 
Physics news on Phys.org
kasse said:
Since each photoelectron has energy 1.1E-19, my estimated number of photoelectrons is 4E-6/1.1E-19 = 3.6E13 photoelectrons.

My book says 1.0E13. What have I done wrong?
You divided by the KE of the ejected electrons instead of the energy of the incident photons.
 

Similar threads

  • · Replies 35 ·
2
Replies
35
Views
4K
  • · Replies 25 ·
Replies
25
Views
5K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K