But where ever I read, thermal voltage is provided as 26 mV however it needs to be 26 milli electron volts, right?
Can you you re welcome clarify this?
Yes. A volt is also a Joule per Coulomb. A truth I never ever forget because that a moment. Think of it this way:
If you have actually two bowl separated through some void in a vacuum in external space, and place 100 volts across them, and then imagine place one Coulomb of charge at the negative plate, that Coulomb that charge will be increased towards the hopeful plate. Eventually, the charge will certainly strike the confident plate and also the impact energy will be simply 100 Joules. (And so the must also be the their velocity at affect is the same, so that the kinetic identical \$\frac12 m v^2\$ works out, correctly.)
You have the right to move the plates additional apart and the acceleration will certainly be less, but the street will also be just enough further personal to permit that Coulomb of charge to have exactly the same impact energy. (Because it will have exactly the same velocity in the end and therefore the very same kinetic energy.) that doesn"t issue how much apart, or just how close the bowl are, the results are always the same: 100 Joules of kinetic (impact) energy converted indigenous the front 100 Joules that potential energy. The only difference is the time it bring away for the conversion.
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It"s consistent. 1 volt will certainly impart 1 Joule top top 1 Coulomb in this situation.
Since girlfriend are into units and also dimensional analysis (and i think the is a very good thing), girlfriend might additionally now look into magnetics and also the concept of Webers (or Joule-seconds per Coulomb or Volt-seconds.) A Joule-second is a unit of angular momentum. Electrons have actually quantized "spin," and they also have orbitals about an atom and also there is something there to consider about angular momentum.
In fact, electron spin and also orbital movement (or, in ~ least, under the presumptions of models because that these) both add to magnetic dipole moment. In magnetic materials, these tiny moments do not include up come zero.
In every these situations, they have angular momentum and the magnetic dipole moment will it is in proportional come the angular momentum, by part proportionality factor.
So, supposing this relationship and also using \$\mu\$ for the magnetic dipole moment and \$L\$ together the angular momentum, we have (using \$x\$ for currently as the factor):
$$\mu = x\cdot L$$
For one electron in a one orbit (Bohr"s early idea because that atoms) and also using \$\vecp\$ together the momentum vector that the electron and also \$\vecr\$ as the vector native the facility of the atom:
$$\mid \vecL\mid\: =\: \mid \vecr \times \vecp\:\mid \: =$$
But together this is circular and using \$R\$ as the scalar for the orbit radius,
$$L = R\: p\:\operatornamesin\left(90^\circ\right) = R\: m_e\: v_e$$
For a present loop, and also you deserve to think of one electron in a circular orbit together being simply such a present loop, we understand that \$\mu=I\left(\pi R^2\right)\$. Also, the time it take away to make an orbit is \$T=\frac2\pi Rv_e\$ and also of course the electrical charge of one electron is \$q_e\$, therefore \$I=\fracq_e v_e2 \pi R\$ (charge every unit time.)
Therefore the magnetic dipole moment of a single, circular orbiting electron is:
$$\mu=I\left(\pi R^2\right)=\fracq_e v_e2 \pi R\left(\pi R^2\right)=\frac12 q_e R v_e$$
(Assuming the \$v_e \ll c \$.) It currently follows:
$$\mu = \fracm_em_e \frac12 q_e R v_e= \frac12 \fracq_em_e R \:m_e\: v_e= \frac12 \fracq_em_e L$$
Clearly then, the lacking factor should be \$x=\frac12 \fracq_em_e\$!!
Angular momentum is quantized. (The change in physics native the at an early stage 1900"s.) Therefore, \$L=N \hbar\$ (where \$\hbar=\frach2\pi\$.) \$N\$ is any integer native 0, up. Assuming \$N=1\$, \$L=\hbar\$. Therefore it adheres to that \$\mu=\frac12\fracq_em_e\hbar\$. For a solitary electron, this functions out come \$\mu_e\approx 1\times 10^-23\:\textrmA\cdot\textrmm^2\$.
Suppose you have actually a bar magnet weighing \$69.5\:\textrmg\$ and that girlfriend go with a process of measure its dipole magnetic moment using a compass and also a yardstick and also measuring the deflection family member to the local magnetic field of the Earth. You uncover that this is \$3.5\:\textrmA\cdot\textrmm^2\$.
Now mean that you assume the almost every one of the atom in this magnet are iron atoms. The massive of one mole of steel is around \$56\:\textrmg\$. So the number of atoms here, using Avagadro"s number, would certainly be \$n=\frac69.5\:\textrmg56\:\textrmg\cdot 6.02\times 10^23 \approx 7.5\times 10^23\$. So currently you"d compute:
$$\mu = n\: \mu_atom = 7.5\times 10^23\cdot 1\times 10^-23\:\textrmA\cdot\textrmm^2 = 7.5\: \textrmA\cdot\textrmm^2$$
This is in reality pretty close come the bar magnet measurement. The amazing thing here is that there are additionally some small contributions come the spin of an electron (ignored here) and also assumptions that every one of the atoms contribute exactly one together electron moment. Also, we assumed one orbits. Bohr"s design was modified by Sommerfeld to very first include elliptical orbits come explain specific hyperfine transitions and later modified to encompass motion in 3 dimensions by Sommerfeld, quiet later. And also then it to be again modified once Uhlenbeck and also Gaudsmit included the ide of spin. (Pauli then come up v his exclusion principle based on all the quantum numbers by then.) but it transforms out in practice that you can really carry out some amazing predictions around such behavior using simplified model approximations.
Note the I had to introduce quantum concept and the straightforward Bohr atomic model to gain here. It"s things prefer this the helped pressure physicists right into developing and also accepting quantum theory.
I guess: v I just want to contact your attention to just how much deserve to be done through dimensional evaluation and thinking about the units in various ways (not the one obvious way, but by changing things out and combining units in different ways that you"d otherwise expect, utilizing a most imagination.) the takes you areas you might not otherwise go and also you might discover brand-new ways that looking at the world approximately you!
Now look at the Ohm, i m sorry is a Joule-second every Coulomb^2. Where can your creativity go through that?
I recommend you check out a book called issue & Interactions, third edition, if interested further.
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Note: 1 Joule = \$6.24150913 \times 10^18\$ eV. You room using \$\fraceVkelvin\$ for your Boltzmann"s consistent rather than \$\fracvoltkelvin\$. Therefore the value of q, the charge on one electron, have to be offered consistently. If you usage \$\fraceVkelvin\$ because that the systems of Boltzmann"s constant, then you must q=1 because that the unit of charge (eV is in very tiny units associated with the smallest unit the charge, 1 electron.) If you usage \$\fracvoltkelvin\$ because that the devices of Boltzmann"s constant, climate you need to use q in units of Coulombs, instead. Simply keep the size in mind, is all.