Calculating energy/work of a quantum

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SUMMARY

The discussion focuses on calculating the energy of a quantum of light with a wavelength of 590 nm, resulting in an energy value of 2.11 eV. Using the equation E = hf, where h is Planck's constant (6.63 x 10-34 J·s) and f is the frequency calculated as f = v/λ, the energy is determined to be 3.37 x 10-19 J. Since this energy is below the work function of the metal (3.0 eV), the quantum will not produce photoelectrons from the metal surface.

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  • Understanding of Planck's constant (h) and its application in quantum mechanics.
  • Knowledge of the relationship between wavelength, frequency, and the speed of light (v = fλ).
  • Familiarity with the concept of work function in photoelectric effect.
  • Basic proficiency in converting energy units from joules to electronvolts (eV).
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  • Learn about the calculation of energy levels in different quantum systems.
  • Explore the concept of work function in various metals and its relevance in photoemission.
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Homework Statement



1- Calculate the energy, in joules, of a quantum of light with a wavelength of 590nm.
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2- Will this colour be able to produce photoelectrons from the surface of a metal with a work function of 3.0 eV? Explain your reasoning.

2. Homework Equations

E = hf

v = fλ

3. The Attempt at a Solution

v = fλ

f = v/λ

f = 3.0 x108m/s / 5.9 x10-7m

f = 5.08 x1014HzE = hf

E = (6.63 x10-34J)(5.08 x1014Hz)

E = 3.37 x10-19J

Joules to eVs:

3.37 x10-19J / 3.0 x108mThe quantum will produce 3.37 x10-19 J or 2.11 eV of energy.
2-

As this quantum of energy was determined to have an energy value of 2.11 eV, and the work function of the metal is 3.0eV, then no, this colour will not produce photoelectrons as the energy produced is below the threshold of the metal’s surface.

I am confident in my answers to part 1 but I'm not sure if I explained part 2 correctly or not. Any feedback would be great.
 
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Part 2 is adequately explained.
 
kuruman said:
Part 2 is adequately explained.
Ok thank you
 

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