I realize this is kind of a silly question, but if you have tritium water, with the tritium half life of 12.5 years, I expect it would be quite hot. (note, this is not a homework question, I'm just curious).

Lets say you have a liter of water, 1000 grams and each and every oxygen with 2 tritiums on it, atomic weight of 22, so 1000/22 = about 45 moles of lets call it T-water. 6.02x10^23 atoms per mole. 2.7x10^25th t-water molecules, (twice as many tritiums)

So half life means in 12.5 years, 2.7x10^25 tritiums would have kicked off an electron and you'd get OH + 3He + energy. There's about 395 million seconds in 12.5 years, and I realize it wouldn't be uniform, but if we ignore that, 2.7x10^25 / 3.95 * 10 ^ 8, about 6.8 x 10^16 tritium decay's per second.

Now, assuming this chart is right you get 0.0057 MeV per decay.

enter image description here

So, our liter of T-water has a decay energy of about 3.8 x 10^14 MeV per second, or 3.8 x 10^20 electronvolts.

with 6.24 x 10^18th electron-volts per joule, our T-water would generate about 61 joules per second, or 14.5 small calories per second.

That would be enough to warm our liter 1 degree every 1,000/14.5 every 69 seconds.

So, here's my question. Am I missing anything or is my math roughly right. If my math is right, if we assume room temperature to be 20 degrees, it would take a little over 1 hour for T-water to warm from room temperature to boiling temp, but assuming it's not insulated, just in a closed, sealed room, the heat would leave it fast enough that it wouldn't reach boiling temp.

My questions are:

  1. is my math right, or is there an egregious error in there, cause I would think that anything with a 12.5 year half life would be hotter than that.

  2. what would the temperature average out to with 61 joules of heat per liter, assuming in a room where heat conduct and evaporate, but air tight, so you don't lose any of the T-water to the environment. Large enough that the water could evaporate and re-condense freely. Would a lot of the energy be lost in photons/radiation or would it mostly be retained as heat?

  3. what would happen if you stick your hand in the T-water? Would you get radiation burns or would it just feel warm/hot?

  4. Would it glow bright? or emit photons in the Gamma spectrum?

  • 3
    $\begingroup$ As a follow up question, would this be an efficient way to keep my swimming pool heated? $\endgroup$
    – userLTK
    May 16, 2015 at 23:18
  • 8
    $\begingroup$ No, it would not be an efficient way to heat your pool since it costs about 30000 bucks per gram. $\endgroup$
    – hft
    May 17, 2015 at 0:29
  • 2
    $\begingroup$ @CountIblis thermos bottles are only for Ice-nine. $\endgroup$ May 17, 2015 at 11:28
  • 3
    $\begingroup$ @CarlWitthoft Tritium still is used for long time lights (for example emergency exit lighting) and some watch hands. Once upon a time Radium was used for glow in the dark watch hands. $\endgroup$ May 17, 2015 at 14:00
  • 2
    $\begingroup$ Note that tritiated water will boil at a slightly higher temperature than regular water. ($\text{D}_2 \text{O}$ is about 101.4°C, so I'd expect $\text{T}_2 \text{O}$ to be around 102-103°C.) This might help if you want to cook your pasta a little more quickly. $\endgroup$ May 18, 2015 at 14:29

1 Answer 1


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.

  • $\begingroup$ The energy deposition in the water should be cut in half because there is an anti-neutrino produced which does not deposit its energy in the water. A factor 2 does not change the conclusion much. Good answer. $\endgroup$ May 18, 2015 at 3:07
  • $\begingroup$ @RossMillikan thanks - the LBL source does state "beta energy 19 keV", not "total energy". It does use the word "max" in one place but not another. Do you have a source for the factor 2? Does the total energy get split between the two? Always happy to learn... $\endgroup$
    – Floris
    May 18, 2015 at 3:11
  • 1
    $\begingroup$ Yes, the total energy gets split between the electron and anti-neutrino. The nucleus takes up the momentum and is so heavy it doesn't take up significant energy. Wikipedia says the total energy is 18.6 keV and the electron averages 5.7 keV-I had guessed it would get half. $\endgroup$ May 18, 2015 at 3:17
  • $\begingroup$ @RossMillikan - I have updated the calculations. Thanks for your help. $\endgroup$
    – Floris
    May 18, 2015 at 14:09

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