Sorry for naive question: Does a single electron have an atto or femto-amperage?

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Single electron counters can measure the flow of individual electrons, leading to discussions about atto-amperage. The concept of current can be quantified by the charge of an electron divided by its period of motion, suggesting that even a single electron can produce a measurable current. The discussion highlights that current density is influenced by both the number of charges and their drift speed, which can be very low. There is a distinction between theoretical models and practical measurements, with sensitivity of instruments being crucial for detecting small currents. Understanding these principles is essential for grasping the behavior of electrons in various contexts.
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Single electron counters appear to exist, the Jap. have one that even determines direction, and they discuss atto-amperage. What I don't quite get is, if a single electron flows, is there an observed very small current ? Since an electron is the smallest size, more sensitive equipment then can never discover an even smaller current, right ?

I accept the notion that models may call for a smallest indivisible unit, I'm only wondering how we establish that we have indeed located that size. Thanks for commenting..
 
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If the particle undergoes a finite and quasiperiodic motion, then the electric current could be quantified as the charge of the particle divided by the period. For example, for the hydrogen atom:

<br /> \left\{\begin{array}{lcl}<br /> \frac{m_{e} v^{2}}{r} &amp; = &amp; \frac{k_{0} e^{2}}{r^{2}} \\<br /> <br /> m_{e} v r &amp; = &amp; \hbar<br /> \end{array}\right. \Rightarrow <br /> \left\{\begin{array}{lcl}<br /> v^{2} r &amp; = &amp; \frac{k_{0} e^{2}}{m_{e}} \\<br /> <br /> v r &amp; = &amp; \frac{\hbar}{m_{e}}<br /> \end{array}\right.<br />

<br /> \left\{\begin{array}{lcl}<br /> v &amp; = &amp; \frac{k_{0} e^{2}}{\hbar} \\<br /> <br /> r &amp; = &amp; \frac{\hbar^{2}}{m_{e} k_{0} e^{2}}<br /> \end{array}\right.<br />

The period is:

<br /> T = \frac{2 \pi r}{v} = \frac{1}{4 \pi^{2} c} \frac{(h c)^{3}}{m_{e} c^{2} (k_{0} e^{2})^2} = 1.520 \times 10^{-16} \, \mathrm{s}<br />

This gives a current of:

<br /> I = \frac{e}{T} = 1.05 \, \mathrm{mA}<br />
 
thx for formula, could you also paste the terms, definitions ? Also the last term is not clear to me,i.e. the 1.05 mA (!). I'm not in physics, would appreciate discussion of hardware level, if any. I read the wikipedia articles, they did not really center on measurement and instrument sensitivity.
 
No, I can't. Look for Bohr model lectures. mA stands for 'milliampere'. Learn the prefixes for SI units.
 
ndvcx said:
What I don't quite get is, if a single electron flows, is there an observed very small current ?

Current density can be seen as the product of the number of charges and their drift speed:
http://en.wikipedia.org/wiki/Current_density

So according to the first equation in that link, nothing prevents having a single electron going at arbitrary low speed, from which you can easily get an atto-ampere if you want. Current depends on both the number of charges and their drift velocity (itself a function of resistance and voltage in paticular).
 
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