Find minimum velocity of emitted electron

In summary, the photoelectric experiment shows that 5eV of energy is required to remove an electron from platinum, using light with a wavelength of 150nm. To find the velocity of the emitted electron, we can use de Broglie's equation and the excess energy becomes kinetic energy. By converting the work function into energy and solving for frequency, we can find the threshold frequency. Then, using the equation K.E. = h(v-x), where v is the frequency of light, we can solve for velocity and get the answer of 1.058m/s.
  • #1
xiphoid
57
0

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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  • #2
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.
 
  • #3
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.
 
  • #4
xiphoid said:
But I am suppose to find the velocity of the electron?

The excess energy becomes kinetic energy.
 
  • #5
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.
 
  • #6
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.
 

1. What is the "minimum velocity of emitted electron"?

The minimum velocity of emitted electron refers to the minimum amount of kinetic energy required for an electron to be emitted from a material, also known as the work function. This is dependent on the material's properties and can vary for different materials.

2. Why is it important to find the minimum velocity of emitted electron?

Finding the minimum velocity of emitted electron is important in understanding the behavior of electrons in different materials. It also has practical applications in fields such as semiconductor technology and energy conversion.

3. How is the minimum velocity of emitted electron calculated?

The minimum velocity of emitted electron can be calculated using the equation mv2/2 = φ, where m is the mass of the electron and φ is the work function of the material. This equation relates the kinetic energy of the electron to the work function.

4. What factors affect the minimum velocity of emitted electron?

The minimum velocity of emitted electron is affected by the work function of the material, as well as the energy of the incident photon or particle. It can also be affected by external factors such as temperature and electric fields.

5. How is the minimum velocity of emitted electron measured?

The minimum velocity of emitted electron can be measured using various techniques such as photoelectric effect experiments, where the kinetic energy of emitted electrons is measured and used to calculate the work function. Other methods include scanning tunneling microscopy and field emission microscopy.

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