Kinetic Engergy in a Quantum Oscillator

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SUMMARY

The discussion centers on calculating the kinetic energy required for an electron beam to excite a quantum oscillator from its ground state to the second excited state. The mass of the oscillator is specified as 3e-26 kg and the spring stiffness as 80 N/m. The correct kinetic energy value is determined to be 6.8e-2 J, which can be converted to kiloelectronvolts. Participants emphasize the need to equate the electron beam's energy to the energy difference between the oscillator's quantum states.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically energy levels in quantum oscillators.
  • Familiarity with the formula DeltaE = hbar * sqrt(Ks/m) for calculating energy spacing.
  • Knowledge of kinetic energy equations, particularly K + U = (0.5p^2/m) + 0.5Ks^2 + U0.
  • Basic skills in unit conversion, especially between joules and kiloelectronvolts.
NEXT STEPS
  • Study the derivation of energy levels in quantum harmonic oscillators.
  • Learn about the relationship between kinetic energy and potential energy in quantum systems.
  • Explore the conversion methods between joules and kiloelectronvolts for better understanding of energy units.
  • Investigate the implications of spring stiffness (K) on the energy levels of quantum oscillators.
USEFUL FOR

Students and professionals in physics, particularly those focusing on quantum mechanics, as well as engineers working with quantum oscillators and electron beams.

cowmoo32
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How much kinetic engergy in eV must an election beam have to be able to excite a quantum oscillator from its ground state to two levels above the ground state if the mass is 3e-26kg and the spring stiffness is 80N/m?

All I can find is the spacing between the energy levels, I have no idea how to find K. The answer is 6.8e-2 J, but I don't have a clue how to get there.
 
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I can only assume that you can convert whatever energy (number) you got to kiloelectronvolts.

Daniel.
 
I got 2.8e61 using the formula DeltaE = hbar*sqrt(Ks/m)
but this is the space between engergy levels. I don't know how to calculate the kinetic energy of the electron beam.
 
my book has the formula K + U = (.5p^2/m) + .5Ks^2+U0
but I'm not sure what K would be equal to. Granted, it's obvious it would be (.5p^2/m) + .5Ks^2+U0-U but I don't know U either.
 
It means that you don't understand the question: the electron beam's energy is at least equal to the energy the quantum oscillator gets in order to jump from the fundamental level to the second excited one. For simplicity, take it as equal.

Daniel.
 
Ok, the beam has to have as much energy as it takes for the electron to move from one energy level to the next, I understand that. But I still don't understand how to arrive at the answer. The answer I got is nowhere near the correct answer.
 
Maybe it's a number/conversion problem. I'm n ot going to show such trivial computation and maybe no one will.

Daniel.
 

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