Calculating Time of Excited Electron in Ground State

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What determines the amount of time that an electron spends in an excited state before it drops back to its ground state?

Is the attraction between the electron and the nucleus the cause? Would the electrons in heavy elements spend less time in their excited states?

I'm conceptualising an excited electron as similar to a ball thrown in the air (that must fall back to earth).

If we:

1. (erroneously) assume that an electron can travel in the space bewteen energy levels; and
2. apply classical mechanics to predict how long the electron should take to reach its peak before falling to its ground state (due to attractive force of nucleus),

would that time be the same as the time it takes for the electron to make a quantum leap to its excited state and then back to its ground state?


Thanks
 
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well delta E, delta t(time) >_ Plancks constant
there you can calculate the time if you know the energy where
E= (pi)^2 (plancks)^2 (n)^2/(2mL^2)
where n is the energy level and L is the diameter of an electron.
There you can calcuate the time it will spend on the energy state.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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