What Are the Effects of Light Wavelength on Photoelectron Production?

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SUMMARY

The discussion focuses on the effects of light wavelength on photoelectron production, specifically using a photocathode with a work function of 2.9 eV illuminated by white light from 400 nm to 700 nm. The threshold frequency for photoelectron production is calculated as 7 x 1014 Hz. The correct wavelength range for photoelectron production is determined to be 400 nm to 428 nm, and the deBroglie wavelength of an emitted electron with 0.2 eV kinetic energy is calculated to be 2.7 nm. Accuracy in calculations and significant figures are emphasized as critical factors affecting the results.

PREREQUISITES
  • Understanding of photoelectric effect and work function
  • Familiarity with the equations for frequency and wavelength (f0 = E0/h, c = fλ)
  • Knowledge of deBroglie wavelength calculations (λ = h/mv)
  • Basic principles of significant figures in scientific calculations
NEXT STEPS
  • Study the photoelectric effect and its applications in modern technology
  • Learn about the significance of work function in different materials
  • Explore advanced calculations involving kinetic energy and deBroglie wavelength
  • Research the impact of light wavelength on electron emission in various photocathodes
USEFUL FOR

Students in physics, educators teaching the photoelectric effect, and researchers interested in electron behavior under varying light conditions will benefit from this discussion.

kevnm67
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Homework Statement


A photocathode whose work function is 2.9 eV is illuminated with white light that has a continuous wavelength band from 400 nm to 700 nm.
A) What is the threshold frequency for photoelectron production?
a. 4 x 1014 Hz b. 7 x 1014 Hz c. 13 x 1014 Hz
B) What is the range of the wavelength band (in nm) in this white light illumination for which photoelectrons are produced?
a. 428 nm to 700 nm b. 400 nm to 428nm c. 300 nm to 400 nm
C) What is the deBroglie wavelength of an emitted electron with kinetic energy of 0.2 eV?
a. 2.5 nm b. 1.2 nm c. 2.7 nm



Homework Equations



f0 = E0/h

l = h/mv

c= f l


The Attempt at a Solution



I want to make sure I am solving these questions correctly. Here is what I did:

a) f0 = 2.9eV/4.14x10-15 = 7 x 1014

b) c = f l
=429 nm
So I said the range would start at 400 and go up to 429nm...the answer was 400-428 nm

c) l = h/mv
m= 9.11 x 10-31
v= 2.65 x 105 which I got from using 1/2 mv2=.2eV
I multiplied .2 eV by 1.6 x 10-19 J to get K
then squrt (2 x K/mass of electron)

l = (6.63 x 10-34J s)/(9.11 x 10-31)(2.65 x 105 m/s) = 3 nm

The answer was 2.7 nm

I wasn't sure if I was rounding differently or maybe did something wrong? Thanks for the help!
 
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I can see why some of the answers differs from the correct ones. In b), since you like to use rounded values for your calculations, I assume you used 3*108 m/s for the speed of light in vacuum? Dividing this with the given answer from a will make your answer deviate. It all comes down to the accuracy of the values you used in your calculations. [3*108]/[7*1014] m deviates by 0.13 % whereas [2.998*108]/[7*1014] m deviates by 0.07 %, [2.99792458*108]/[7*1014] m by 0.06 %. Of course, it also depends on what level of accuracy the makers of the question used as [4.135669*10-15*2.99792458*108]/[2.9] m deviates by 0.11 % but is lucky enough to be rounded to the correct value.

In c it's basically the same thing - accuracy. Although, here there's also the number of significant figures to consider. You've rounded to one significant figure, which would be correct to do since the given information for the calculations gives the least number of significant figures as one. So based on that, I would say 3 nm is allowed. However, since you're given multiple answers to choose from, you'd be better of by choosing the one that comes the closest to the answer in the calculations.
 

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