Atomic clock: energy between two levels

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SUMMARY

The atomic clock defines a second based on the oscillation of electromagnetic radiation at a frequency of 9,192,631,770 Hz, corresponding to the energy gap between two levels of a caesium-133 atom. The energy difference can be calculated using the equation E=hf, where h is Planck's constant (6.626 x 10-34 Js). The resulting energy difference is approximately 3.8 x 10-5 eV. The calculations and assumptions made in the discussion align with established physics principles and can be verified through resources like HyperPhysics.

PREREQUISITES
  • Understanding of electromagnetic radiation frequency and oscillation periods
  • Familiarity with Planck's constant (6.626 x 10-34 Js)
  • Knowledge of the relationship between energy, frequency, and Planck's equation (E=hf)
  • Basic proficiency in unit conversion between joules and electronvolts (eV)
NEXT STEPS
  • Explore the concept of atomic clocks and their applications in timekeeping
  • Learn about the principles of quantum mechanics related to energy levels in atoms
  • Investigate the significance of caesium-133 in defining the second
  • Study the implications of frequency and energy calculations in modern physics
USEFUL FOR

Students in physics, engineers working with precision timing devices, and anyone interested in the fundamental principles of atomic clocks and quantum mechanics.

Jillds
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Homework Statement


With the atomic clock a second is defined as the time it takes for EM radiation to oscillate 9192631770 times, which equals the energy gap between two energy levels of a caesium-133 atom. Note: it's a translation and the term used with the oscillation is "oscillation periods"

Homework Equations


Calculate the energy difference (in eV)

The Attempt at a Solution


[/B]
At first I took the expression "oscillation periods" to mean that I had to use the T=1/f relation and insert it in the E=hf equation. However since T is in seconds and f means turns or oscillations, I went with f=9192631770.

So, I have:
E = hf
= 6.626*10^(-34) Js * 9192631770 /s
= 6.091*10^(-24) J
= 3.8 * 10^(-5) eV

I just hope to check whether I made the right assumptions and conversions, thank you.
 
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Jillds said:

Homework Statement


With the atomic clock a second is defined as the time it takes for EM radiation to oscillate 9192631770 times, which equals the energy gap between two energy levels of a caesium-133 atom. Note: it's a translation and the term used with the oscillation is "oscillation periods"

Homework Equations


Calculate the energy difference (in eV)

The Attempt at a Solution


[/B]
At first I took the expression "oscillation periods" to mean that I had to use the T=1/f relation and insert it in the E=hf equation. However since T is in seconds and f means turns or oscillations, I went with f=9192631770.

So, I have:
E = hf
= 6.626*10^(-34) Js * 9192631770 /s
= 6.091*10^(-24) J
= 3.8 * 10^(-5) eV

I just hope to check whether I made the right assumptions and conversions, thank you.
Your answer matches the one here: http://hyperphysics.phy-astr.gsu.edu/hbase/acloc.html
 
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