Basic quantum physics, Balmer series

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SUMMARY

The discussion focuses on calculating the wavelengths of the first three ionization energies of hydrogen in the Balmer series. The ionization energies provided are 1312, 328, and 146 kJ/mol. The calculations involve using the Planck constant (h = 6.626 x 10-34 Js) and the speed of light (c = 2.998 x 108 m/s) to derive the energy of one photon and subsequently the wavelength. It is concluded that the calculated wavelength of 9.1178 x 10-5 nm is incorrect, as it should fall within the visible spectrum (400-780 nm).

PREREQUISITES
  • Understanding of ionization energies and their significance in quantum physics
  • Familiarity with the Balmer series and its definition
  • Knowledge of the Planck constant and its application in energy calculations
  • Ability to convert between energy units, specifically from kJ/mol to Joules
NEXT STEPS
  • Learn about the Rydberg formula for calculating wavelengths in the hydrogen spectrum
  • Study the relationship between ionization energies and electron states in hydrogen
  • Explore the concept of electron-volts and their conversion to Joules for easier calculations
  • Investigate the significance of the Balmer series in atomic physics and its applications
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Students studying quantum physics, particularly those focusing on atomic structure and spectral lines, as well as educators teaching the principles of the Balmer series and ionization energies.

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Homework Statement


The first five ionization energies of hydrogen are 1312, 328, 146, 82 and 52 kJ/mol. Calculate the wawelength of the first three in Balmer series.


Homework Equations


Avokadro constant, NA = 6,022 x 1023 /mol
Planck constant, h = 6,626 x 10-34Js
speed of light, c = 2,998 x 108ms-1
energy, E in J
wavelength, lambda in nm = 10-9m
frequency of radiation, v in s-1

E = hv
c = lambda x v

The Attempt at a Solution


So, in my knowledge you only need the first three energies given in problem and the energies represents the value of one mole of photons. So dividing the 1312 with Avokadro you will get the energy of one photon which is 2.17868 x 10-21J

Using the equation E = hv you will get v = 3,28807 x 1012s-1
and using c = lambda x v you will get lambda = 9,1178 x 10-5nm

So the exponent is definitely wrong but so are numbers because this wavelength should be visible to human eye (400-780nm). You can solve this other way when you are not using the given energies and the rydberg value is present in the equation but this should be another way to do this.
Thx beforehand
 
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One suggestion I have is that you get comfortable working with the electron-volt as a unit of energy. The conversion is 1 eV = 1.6x10-19 J. It makes the numbers quite a bit easier to deal with.

Here's another trick that can save you some time. The combination hc is approximately 1240 nm eV, so the energy and wavelength of a photon are related by[tex]E = \frac{hc}{\lambda} = \frac{1240~\mathrm{nm~eV}}{\lambda}[/tex]

Regarding the actual problem, you want to look what the definition of the Balmer series is. Also, how are the ionization energies related to the energy of the electron states?
 

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