Find minimum velocity of emitted electron

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Discussion Overview

The discussion revolves around calculating the minimum velocity of an emitted electron in a photoelectric experiment involving platinum, where 5 eV of energy is required to remove an electron and light of 150 nm wavelength is used. The conversation includes attempts to apply relevant equations and concepts to find the velocity of the emitted electron.

Discussion Character

  • Homework-related, Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant notes the requirement of 5 eV to remove an electron and suggests converting the 150 nm wavelength to photon energy to find the energy difference that gives the electron's energy.
  • Another participant claims to have calculated the electron's energy as 77 eV but questions its correctness while emphasizing the need to find the velocity.
  • It is mentioned that the excess energy from the photon becomes the kinetic energy of the emitted electron.
  • A participant proposes a method involving converting the work function into energy, calculating the threshold frequency, and then using kinetic energy equations to find the velocity.
  • One participant expresses confusion about their calculated velocity of 1.058 m/s, indicating a potential mistake and seeking feedback on their approach.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct velocity or the calculations involved, with multiple competing views and methods presented without resolution.

Contextual Notes

Participants reference the need to convert units and apply various equations, but there are unresolved steps in the calculations and assumptions regarding energy conversions and definitions.

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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