Calculate the minimum frequency needed to start the electron

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SUMMARY

The minimum frequency required to extract an electron from a metal with a work function of 3.8 eV is calculated using the photoelectric effect equation, \(h\nu = h\nu_0 + \frac{1}{2}mv^2\). The calculation initially yielded a frequency of \(5.73 \times 10^{33}\) Hertz, but the user was advised to ensure unit consistency. Specifically, the energy should be converted from electronvolts (eV) to joules (J) or Planck's constant (h) should be used in eV-sec. The correct conversion factor is \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\).

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with Planck's constant (h)
  • Knowledge of energy units (eV and J)
  • Basic skills in dimensional analysis
NEXT STEPS
  • Convert energy values between electronvolts and joules
  • Study the implications of the photoelectric effect in quantum mechanics
  • Learn about the significance of Planck's constant in physics
  • Explore dimensional analysis techniques for unit consistency
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Students in physics, educators teaching the photoelectric effect, and anyone interested in quantum mechanics and energy calculations.

pyxolo
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Hi there.
I am a student from Spain and I am about to join the University but I have to pass physics at High School.

Here It is:

The work of extracting an Eletron in a metal is 3.8 eV. Calculate the minimum frequency needed to start the electron.


This is what I did.

[tex]\[E = \phi + Ec\][/tex]

Regarding to Einstein's equation of photoelectric effect:

[tex]\[h\nu = h\nu _0 + \frac{1}{2}mv^2 \][/tex]

When I got to this point, I assumed that the electron's velocity would be zero and then:

[tex]\[\nu = \frac{{3.8}}{{6.63\cdot10^{ - 34} }} = 5.73\cdot10^{33}[/tex]
Hertzios

Is everything alright?
 
Last edited:
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No. You have to be more careful with units. You have used the energy in units of eV, and h in units of J-sec. You need to use a consistent set of units - either convert the energy in eV into J, or use h in eV-sec. Try again, and use dimensional analysis to check the units of your answer.
 
lol, I didn't notice, [tex]\[<br /> 1eV = 1.602\cdot10^{ - 19} J<br /> \][/tex]
 

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