Calculate the frequency, in hertz (level n = 10 to level n = 5)

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SUMMARY

The discussion focuses on calculating the frequency of a photon emitted when an electron transitions from energy level n = 10 to n = 5 in a hydrogen atom. The Rydberg formula, 1/λ = RH ∙ (1/n₁² - 1/n₂²), was correctly applied, yielding a wavelength of λ = 3.0392×10⁻⁶ m. However, the user miscalculated the frequency, converting the wavelength to Hz incorrectly, resulting in an answer of 9.89e14 Hz. The importance of considering the atomic number (Z) for hydrogen-like atoms was also highlighted, indicating that the Rydberg formula requires adjustment for different atomic numbers.

PREREQUISITES
  • Understanding of the Rydberg formula for hydrogen: 1/λ = RH ∙ (1/n₁² - 1/n₂²)
  • Knowledge of photon frequency and wavelength relationship: c = λf
  • Familiarity with atomic structure and energy levels in hydrogen atoms
  • Basic understanding of atomic number (Z) and its effect on spectral lines
NEXT STEPS
  • Research the Rydberg constant (RH) and its application in spectral calculations
  • Learn about the impact of atomic number (Z) on the Rydberg formula for hydrogen-like atoms
  • Study the relationship between wavelength, frequency, and the speed of light (c = λf)
  • Explore examples of photon emission calculations for different transitions in hydrogen and hydrogen-like atoms
USEFUL FOR

Students and educators in physics, particularly those studying atomic physics and quantum mechanics, as well as anyone interested in understanding photon emission and spectral analysis in hydrogen atoms.

pinklemon
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Calculate the frequency, in hertz, of the photon that is emitted when a electron falls from level n = 10 to level n = 5 in the H atom.

Enter a numeric answer only, no units.


I tried asking in other places but ppl are so ambiguous about it.
:( please help me.

I used rydberg formula:
1/λ = RH ∙ (1/n₁² - 1/n₂²)
(RH rydberg constant for hydrogen, n₁ lower level, n₂ higher level)

Hence:
λ = 1/ [ 1.09678×10⁷m⁻¹ ∙ (1/5² - 1/10²) ] = 3.0392×10⁻⁶m

and then I converted to Hz = 9.89e14 but i still GOT IT wrong :( Please help me someone!
 
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The above calculation is done for a hydrogen atom. Check the problem to see if a different Z is given. (the factor of Z^2is added to the rydberg equation for approximating photon frequencies of hydrogen-like atoms with a different atomic number.
 
elduderino said:
The above calculation is done for a hydrogen atom. Check the problem to see if a different Z is given. (the factor of Z^2is added to the rydberg equation for approximating photon frequencies of hydrogen-like atoms with a different atomic number.

I don't understand. So there's another Z? But I thought it is only 5 and 10?

oh dear. i think i posted in the wrong section.
(please move this thread) thanks!
 
Last edited:

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