How do you work out the wavelegth (Franck-Hertz experiment)?

Click For Summary
SUMMARY

The Franck-Hertz experiment with potassium vapor indicates that the current drops sharply at an applied voltage of 1.62V, which corresponds to an energy change (delta E) of 1.62 eV. The wavelength of the expected spectral line in the emission spectrum can be calculated using the formula λ = hc/E, resulting in a wavelength of 767 nm when using the correct constants. The value 1.24 is derived from the conversion factor used in the equation λ = 1.24/eV, which is essential for accurate wavelength calculations. For precise results, it is crucial to verify the values and units of the constants from reliable sources such as the NIST Reference on Constants, Units, and Uncertainty.

PREREQUISITES
  • Understanding of the Franck-Hertz experiment and its significance in quantum mechanics.
  • Familiarity with the equation E = hf = hc/λ for energy-wavelength calculations.
  • Knowledge of the Planck constant and the speed of light in vacuum.
  • Ability to convert energy in electronvolts (eV) to wavelength in nanometers (nm).
NEXT STEPS
  • Review the NIST Reference on Constants, Units, and Uncertainty for accurate physical constants.
  • Learn about the significance of the Franck-Hertz experiment in atomic physics.
  • Study the relationship between energy levels and spectral lines in atomic emission spectra.
  • Explore the derivation and application of the equation λ = hc/E in various contexts.
USEFUL FOR

Students studying quantum mechanics, physics educators, and researchers interested in atomic spectroscopy and energy transitions in atoms.

birdnotbeing
Messages
1
Reaction score
0

Homework Statement


In a Franck-Hertz experiment carried out with potassium vapour, it is found that the current falls off rapidly at an applied voltage of 1.62V. Calculate the wavelength of the expected spectral line in the emission spectrum of potassium when this voltage is reached.

Homework Equations


E = hf = hc/lambda

The Attempt at a Solution


delta E = 1.62 eV
lambda = hc/1.62 = 3.08x10^-25

According to a powerpoint, it should be 1.24/eV = 1.24/1.62 = 767nm (which is right). Where does 1.24 come from?
 
Physics news on Phys.org

Similar threads

Replies
5
Views
7K
Replies
5
Views
7K
Replies
1
Views
40K