Find minimum velocity of emitted electron

AI Thread Summary
To find the velocity of the emitted electron in a photoelectric experiment using light of 150 nm wavelength, the energy of the photon must first be calculated. The energy required to remove an electron from platinum is 5 eV, and the energy of the photon can be derived from its wavelength. The difference between the photon energy and the work function gives the kinetic energy of the emitted electron. By applying the kinetic energy formula and solving for velocity, the correct answer is determined to be approximately 1.058 m/s. The discussion emphasizes the importance of accurately converting units and applying the relevant equations to arrive at the solution.
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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