Question:
What is the required frequency stir an atom is a trap of 20 micron radius to put the an atom with in $ L_3 = 1 \times \hbar $?
Solution:
Disclaimer - I have got this problem wrong - and there is no official solution yet, so this is at best a sharing of my idea.
The Bohr model gives $ mvr = n\hbar $. The problem requires $ n = 1 $, so we got almost everything in this equation, except $ v $ .
If a stirrer is operating at a certain frequency $ f $, then a particle being stirred will have travelled a circle in one period of time, in other words, $ v = \frac{d}{T} = \frac{2\pi r}{T} = 2\pi r f $.
Substitute this back to the equation we get
$ m (2 \pi r f) r = \hbar $. So $ f = \frac{\hbar}{2 \pi r^2 m} = \frac{h}{ r^2 m} = 43 Hz $
I didn't know what I pick $ 10^6 Hz $ at that point - perhaps I was just too nervous. I am not sure if the current answer is correct either. But that's the idea - the problem isn't too hard. It is just that the climax was just too high :p
What is the required frequency stir an atom is a trap of 20 micron radius to put the an atom with in $ L_3 = 1 \times \hbar $?
Solution:
Disclaimer - I have got this problem wrong - and there is no official solution yet, so this is at best a sharing of my idea.
The Bohr model gives $ mvr = n\hbar $. The problem requires $ n = 1 $, so we got almost everything in this equation, except $ v $ .
If a stirrer is operating at a certain frequency $ f $, then a particle being stirred will have travelled a circle in one period of time, in other words, $ v = \frac{d}{T} = \frac{2\pi r}{T} = 2\pi r f $.
Substitute this back to the equation we get
$ m (2 \pi r f) r = \hbar $. So $ f = \frac{\hbar}{2 \pi r^2 m} = \frac{h}{ r^2 m} = 43 Hz $
I didn't know what I pick $ 10^6 Hz $ at that point - perhaps I was just too nervous. I am not sure if the current answer is correct either. But that's the idea - the problem isn't too hard. It is just that the climax was just too high :p
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