Find minimum velocity of emitted electron

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SUMMARY

The discussion focuses on calculating the minimum velocity of an emitted electron from platinum when exposed to light of 150 nm wavelength, requiring 5 eV of energy to remove the electron. The participants utilize de Broglie's equation and the relationship between photon energy and kinetic energy to derive the electron's velocity. The correct approach involves converting the work function (5 eV) into energy, determining the threshold frequency, and applying the kinetic energy formula K.E. = h(v - x) to find the velocity. The final calculated velocity of the emitted electron is 1.058 m/s.

PREREQUISITES
  • Understanding of the photoelectric effect and work function
  • Familiarity with de Broglie's equation
  • Knowledge of energy conversion from wavelength to photon energy
  • Ability to apply kinetic energy formulas in physics
NEXT STEPS
  • Learn about the photoelectric effect and its implications in quantum mechanics
  • Study the derivation and applications of de Broglie's equation
  • Explore the conversion of wavelength to energy using E = hc/λ
  • Investigate kinetic energy calculations in particle physics
USEFUL FOR

Students in physics, particularly those studying quantum mechanics, as well as educators and anyone interested in the practical applications of the photoelectric effect and electron dynamics.

xiphoid
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Homework Statement


Photoelectric experiment show that about 5eV of energy are required to remove an electron from platinum, if light of 150nm wavelength is used, what is the velocity of the emitted electron?

Homework Equations


de broglie's equation, E=hv

The Attempt at a Solution


Tried by using the de Broglie's equation, to find the velocity but wasn't successful.
 
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5eV is energy, and 150 nm can also be converted to the energy of the photon. The diff between these two gives you the energy of the electron.
 
The energy of the electron, i got is 77eV, is it correct?
But I am suppose to find the velocity of the electron?

voko said:
5eV is energy, and 150 nm can also be converted to the energy of the photon. The diff between these two gives you the energy of the electron.
 
xiphoid said:
But I am suppose to find the velocity of the electron?

The excess energy becomes kinetic energy.
 
The answer what i got after considering 77eV as the kinetic energy, was not the original one, the answer should be 1.058m/s
I mentioned in my previous comments all the possible values that I managed to find, kindly check them and let me know if i have made a mistake, even the concept and its application seems to be right to me as per the hints received.
voko said:
The excess energy becomes kinetic energy.
 
Hi ill try to explain it.
Convert the workfunction ( 5ev) into energy and then solve
E=hv for v ( frequency) youll get the threshold frequency ( we call it x) that is the frequency required. then
put the value in the following equation
K.E.=h(v-x) v is the frequency of the light and then solve it for velocity and youll get the answer.
 

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