Solve Ionization & Light: Frequency for Hydrogen

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SUMMARY

The frequency of light required for the ionization of hydrogen, as derived from the Bohr model, is 3.29 × 1015 Hz. The calculation involves using the formula ΔE = hν, where ΔE is the ionization energy, h is Planck's constant (6.63 × 10-34 J·s), and ν is the frequency. The first ionization energy of hydrogen is 1312 kJ/mol, which converts to 2.1783 × 10-18 J per electron when divided by Avogadro's number (6.023 × 1023 mol-1).

PREREQUISITES
  • Understanding of the Bohr model of the hydrogen atom
  • Familiarity with Planck's constant (h = 6.63 × 10-34 J·s)
  • Knowledge of Avogadro's number (6.023 × 1023 mol-1)
  • Basic grasp of energy-frequency relationships in quantum mechanics
NEXT STEPS
  • Research the derivation of the Bohr model for hydrogen
  • Learn about the concept of ionization energy in different elements
  • Explore the relationship between wavelength and frequency using ν = c/λ
  • Study the implications of quantum mechanics on atomic structure and behavior
USEFUL FOR

Students studying quantum mechanics, physics educators, and anyone interested in the principles of atomic ionization and light interaction with matter.

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[SOLVED] Ionization and light

I've been working on this problem and I'm not coming up with a reasonable answer.

If the Bohr model is used, what frequency of light
would be required for ionization of hydrogen?

I know that [tex]\Delta E = h \nu[/tex], where E is energy, h is Planck's constant, and [itex]\nu[/itex] is frequency and that [tex]\nu = \frac{c}{\lambda}[/tex] and I know the first ionization energy of hydrogen is 1312 kJ/mol, but I can't seem to arrive at the answer.

I tried using both 1312kJ and 1.312x10^6J for E but the answer comes out too large. The solution is 3.29 × 10^15 Hz. Any help is appreciated.
 
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First step: Questions like this should be on the homework forum, so you probably might want to post there next time.

Next step: Divide the ionization energy (in J) per mol by 6.023x10^23 to get the ionization energy per electron. You should get a number like 2.1783x10^-18 J which should look somewhat familiar to you. This is your E.

Next Next step:

2.7183X10^-18 J = (6.63x10^-34 J*s)f

f = 3.29x10^15 Hz
 
Yep, this is actually my first homework question and I forgot about that forum, actually. Thanks for your response!
 

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