Answer: Maximum Work Function for 900nm IR Light

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SUMMARY

The maximum work function for a surface to emit electrons when illuminated with 900nm infrared light is calculated to be 2.21 x 10-19 J. This value is derived from the photoelectric effect equation, where the energy of the incident light (hf) equals the work function when the kinetic energy (K) of the emitted electrons is zero. The frequency of the light is determined to be 3.33 x 1014 Hz using the speed of light and the wavelength. The calculations confirm that the work function must be sufficient to release electrons without imparting kinetic energy.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with Planck's constant (h = 6.626 x 10-34 J·s)
  • Knowledge of the relationship between wavelength, frequency, and the speed of light
  • Basic principles of kinetic energy (K = 1/2 mv2)
NEXT STEPS
  • Study the photoelectric effect in detail, focusing on the equation hf = K + work function
  • Learn about the implications of work function in different materials
  • Explore the relationship between wavelength and frequency in electromagnetic radiation
  • Investigate the applications of the photoelectric effect in modern technology
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Students in physics, particularly those studying quantum mechanics and the photoelectric effect, as well as educators and professionals involved in materials science and photonics.

babypudding
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Maximum work function??

Homework Statement


What is the maximum work function for a surface to emit electrons when illuminated with 900nm infrared light?


Homework Equations


K= 1/2mv2
= hf - work function

wavelength =h/p = h/mv

The Attempt at a Solution



p=h/wavelength
= 6.626 x 10-34 / 900 x 10-9
= 7.362 x 10-28 kgm/s

p=mv
v=p/m
=7.362 x 10-28 / 9.11x10-31
= 1027.66 m/s

K=1/2mv2
=1/2 x 9.11x10-31 x 1027.662
=4.81x10-25 J

frequency x wavelength = c (no sure if this is right, where c is the speed of light)
f= 3x108/ 900 x 10-9
=270

work function= hf- K
= 6.626 x 10-34 x 270 - 4.81x10-25
= -4.81J
My answer looks wrong, so I think I did it wrong. Please lead me to the right direction.
 
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I imagine they're looking for the workfunction to be the energy of the light at that wavelength since that will be enough to free an electron with no kinetic energy.
 


So
K=1/2mv2
=1/2 x 9.11x10-31 x 1027.662
=4.81x10-25 J

That should be right?
 


babypudding said:
So
K=1/2mv2
=1/2 x 9.11x10-31 x 1027.662
=4.81x10-25 J

That should be right?

No. The wavelength of the radiation is 900nm. This is not the wavelength of the electron. So, the value you have for the electron kinetic energy is wrong.

Read Kurdt's advice again:

I imagine they're looking for the workfunction to be the energy of the light at that wavelength since that will be enough to free an electron with no kinetic energy.

If the material had the biggest possible work function it could have and still give off an electron when illuminated with 900nm radiation, then the electrons released will have K=0. Use this kinetic energy in the photoelectric effect equation to solve for what the work function must be in this case:

hf=K-\phi
 


Ok. Shouldn't it be hf = K + work function ?
h= 6.626 x 10-34
K=0
f= 3x108 / 900 x10-9 = 3.33x1014

So hf = 0 + work function
work function = 2.21 x 10-19 J

Is this right now?
 

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