How Is the Magnetic Field Calculated in the Normal Zeeman Effect?

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SUMMARY

The discussion focuses on calculating the magnetic field strength in the context of the Normal Zeeman Effect, specifically for the red line of the Balmer series in hydrogen at a wavelength of 656.5 nm. Participants noted that the observed splitting of 0.065 nm between spectral lines requires conversion of wavelength to energy using the equation E = hc/λ. The magnetic field can then be calculated using the equation E = E0 + μB × B × ml, where μB is the Bohr magneton (9.27400899 × 10^-24 J T^-1). Missteps in directly equating energy and wavelength were highlighted, emphasizing the need for proper unit conversion.

PREREQUISITES
  • Understanding of the Normal Zeeman Effect
  • Familiarity with the Bohr magneton (μB)
  • Knowledge of energy-wavelength conversion (E = hc/λ)
  • Basic principles of quantum mechanics related to atomic spectra
NEXT STEPS
  • Learn about the Normal Zeeman Effect and its implications in atomic physics
  • Study energy-wavelength conversion techniques in quantum mechanics
  • Explore the calculation of magnetic fields in various contexts using μB
  • Investigate the Balmer series and its significance in spectroscopy
USEFUL FOR

Physics students, educators, and researchers interested in atomic spectroscopy, quantum mechanics, and the effects of magnetic fields on spectral lines.

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Homework Statement



The red line of the Balmer series in hydrogen has a
wavelength of 656.5 nm. Suppose that this line is observed to
split into three different spectral lines when placed in a magnetic field, B, due to the Normal Zeeman Effect. What is the
value of the magnetic field if the wavelength splitting between
adjacent lines is 0.065 nm?


Homework Equations



\mu_{B}\ =\ 9.27400899(37)\ \times\ 10^{-24}\ J\ T^{-1}

E=E_0+\mu_B\times B\times m_l

The Attempt at a Solution



I said that E - E0 = .065 nm

and solved for B, saying m_l = 1, but that's not right.

Any ideas?
 
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E is an energy, nm is a length and not an energy. You can convert the wavelength values to energy values, then it will work.
 
Ah right good catch. I tried that and got 3.3*10^8 T, but that was wrong too.

i did E = \frac{hc}{\lambda}

where \lambda is .065 nm

than divided by \mu_B
 
Note that ##E = \frac{hc}{\lambda}## does not imply that ##\Delta E = \frac{hc}{\Delta \lambda}##.
 

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