energyquantum-tunneling
I'm assuming the 'amplitude' is kind of like the MeV it has on it, so could be seen as a product of the voltage applied to that electron in a field.
But how can it 'lose' volts when passing the barrier, but still have the same energy as a whole? Does it increase the frequency in proportion to the voltage decrease?
Shouldn't the energy gain be greater than this formula describes since the energy from the electric field is applied for so long?
The electron gains energy and accelerates until it encounters a collision. This is a statistical process and there's a distribution for the energy loss for many electrons. Then, it can accelerate again from that point on, until it encounters another collision. This process continues. So the time for which the field is applied doesn't matter so much, because the electron is not continuously accelerating. Bursts of acceleration followed by deceleration happen continuously.
To apply your equation correctly at the microscopic level to a single electron, you'd need to look at the force on the electron during the acceleration phase, and subtract from that the energy lost during a collision. You'd then repeat this for as many times as the electron undergoes acceleration/loss from one end of the wire to the other, to obtain the net energy gain.
It's not like there is a net force on the electrons, so it almost seems the concept of impulse needs to be invoked. Granted, I am very confused in general so maybe there is some obvious answer here. I know the voltage actually describes the same thing but how do I calculate it from the electric field and wire length?
Voltage is defined as the line integral over the electric field. In a uniform wire, the electric field is constant and the voltage is just $V = E l$ where $l$ is the wire length. The voltage across the wire in this context describes the net energy change of an electron as it passes from one end of the wire to the other. That energy change is just $q V$.
On the one hand, if I have a field and an object, and the object moves completely from the point where it initially is and where the source of the field is, with no resistance, it will absorb energy all the way along the distance for a fixed quantity of energy (the potential energy), which can be measured as F*d.
Without resistance, yes, it will continue to absorb energy until it reaches the other end, and F*d tells the whole story.
On the other hand, if along the path energy is taken away from the object, exactly enough to balance the accelerational energy (making for a constant velocity), then if the object starts with zero initial velocity it will constantly absorb energy. If I vary the initial velocity from zero on up, the fixed amount of energy absorbed from points a to b will vary from infinity down to ever smaller quantities.
If the electron starts off with zero initial velocity, the net energy it gains traversing the wire is the constant velocity (steady-state velocity) it reaches by the time it gets to the other end.
So, since an electron has a very small net velocity, shouldn't its actual amount of absorbed energy be greater than the energy predicted by the equation F*d (ie W=E*e(m)*l)?
No. The energy the electron would gain in the absence of any resistance is F*d. So, if there are in fact collisions, we'd expect the net outcome to be lower.
updated calculations - based on neutrino energy escaping and vapor inhalation risk
Your math is close but not quite right.
First - the number of tritium atoms.
There are 1000/(16+3+3) = 45 moles (as you said)
This means there are 45*2*$N_A$ = $5.5 \cdot 10^{25}$ atoms of Tritium
Now the half life is 12.3 years or 4500 days, that is $3.9\cdot 10^8 $s.
This means the 1/e time $\tau=t_{1/2}/\ln{2} = 5.6\cdot 10^8 \mathrm{s}$
The number of decays per second is $\frac{5.5\cdot 10^{25}}{5.6\cdot 10^8} = 9.8\cdot 10^{16} s^{-1}$
The energy given off in one decay is 19 keV (source: LBL) - but the mean energy imparted to the electron is less: only 5.7 keV (Wikipedia. The remainder goes to the anti-neutrino that is also produced in the decay. Since the interaction cross section between the neutrino and the water is extremely small, that energy can be considered "lost" in terms of heating the water.
This means the total energy deposited in the water (all of it, since the beta decay has a very short range) is about 90 W
Clearly this is a significant source of heating - about 22 calories per second, so it will heat a liter of water by one degree C in about 50 seconds. In an insulated container, liquid that started at room temperature would boil in about a little over an hour. During that time, it would produce about 14 ml of $^3He$ gas as well (0.6 mmol).
This liquid would probably not glow - the blue glow usually associated with radioactivity is due to Cerenkov radiation (roughly the optical equivalent of a sonic boom) and that requires particles to travel faster than the speed of light in the medium.
An electron with 19 keV is not really relativistic (rest mass 511 keV), so the speed is given near enough by
$$v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2\cdot 19\cdot 10^3 \cdot 1.6 \cdot 10^{-19}}{9.1\cdot 10^{-31}}} \approx 8\cdot 10^7 \mathrm{m/s}$$
This is sufficiently below the Cerenkov limit (which for water with a refractive index of about 1.3 would be about $2\cdot 10^8 \mathrm{m/s}$ - so there will be no blue glow.
But just because it doesn't glow does not mean it's not dangerous.
According to the CDC, a skin dose of 550 Gy would require amputation. Now with half the water near the skin radiating "away" from the skin, we expect the deposited dose per unit mass in the skin to be half of that deposited in the water; this means the dose rate is 45 Gy per second. You would reach the "amputate everything" dose in about 12 seconds. But because the beta has such low energy, it is said to be absorbed in the "dead upper layer of the skin". This might protect you... If it weren't for the fact that if you have tritiated liquid, you have tritiated vapor. Inhalation of that vapor will be really bad for you.
How bad? Assuming that the swimming pool is a comfortable 22 °C, the saturated vapor pressure of water at that temperature is about 20 Torr or 260 Pa. At 60 % relative humidity (not uncommon near a pool) that would be 150 Pa, or about 1/600th of the air. Assuming that all that inhaled tritium is exchanged with the water in the body, and that an average human exchanges 1 m3 of air with the environment per hour (source), this means 1.6 liter of tritium gas per hour - about 70 mmol. This would deposit 0.07 Gy per hour into the body - where there is nothing to protect you. It would take a bit longer to kill you... but kill you it would. Of course swallowing a bit of pool water would really speed up the process, as would any cuts or abrasions. And did you ever notice how crinkly your skin gets when you spend too much time in the water? That's water penetrating the skin. I don't have a good way to estimate that - the internet is full of stories on the subject, but I found no hard data. I suspect that it will be a significant source of tritium entering the body if you swim in your pool.
So even a very short exposure is likely to be most unpleasant. I recommend against using this as a method to heat your swimming pool.
As an aside, there are plenty of instances where Tritium is used as a good "permanent" light source for watch hands, sighting compasses, and other instruments. The short range of the beta particles and the practical half life make it quite good in this role; mixed with an efficient scintillator that turns the energy into visible light, you don't need a lot of tritium to light up a dial at night. You are allowed to use 25 mCi of tritium for this application without needing an NRC license.
Now 25 mCi is 925 MBq - million decays per second. Your liter of tritiated water exceeds this by about 9 billion; and it is in direct contact with the skin (instead of isolated between a few mm of glass which effectively stops the beta radiation).
This is why tritium watches are safe, and your swimming pool is not.
Best Answer
If you have a flux of electrons, the transmitted power is proportional to the number of tunneling electrons. The other electrons are reflected from the barrier. Thus, the barrier just redistributes the incident electron flux into two ones. It does not change the individual electron energy.