# Gamma Ray Decay to potential watts

1. Jun 13, 2012

### ANarwhal

I'm interested in calculating how much power (in Watts) could be produced (assuming a 100% conversion efficiency between Gamma Rays and electrical power) from gamma decay of a 60 Co -> 60 Ni.

How long would this reaction last? Just a few seconds or longer? Are there any decays that last for a long time?

2. Jun 13, 2012

### Simon Bridge

Each gamma decay takes a very short time (order 10^-23 seconds iirc) ... it is a statistical rather than a continuous process.

You can work out the rates from the mean lives of the processes. Imagine you are driving a sensitive waterwheel with raindrops.

3. Jun 13, 2012

### daveb

As Simon mentioned, the actual power will depend upon the activity, which is decays per second. Multiply this by the energy per decay and you have energy per second, which is power. You'll find it takes an enormous activity to generate enough power to light up an electricl light bulb.

4. Jun 13, 2012

### QuantumPion

Cobalt 60 decay emits two 1.3 MeV gammas and a 0.3 MeV beta. This is 4.5e-13 J per decay. To produce 60 W, you would thus need 1.3e14 decays per second. This is equal to 3615 Ci, or about 3 g of Co-60. Co-60 has a half-life of 5.2 years which means if the bulb initially produced 60 W, after 5.2 years it would produce 30 W, after 10.4 years - 15 W, etc.

Note that this 60 W cobalt-60 lightbulb would cause a dose of 500 R/hr (fatal) at a distance of 10 ft if unshielded. However with a 5 inch thick leaded glass shield surrounding the bulb, this could be reduced to 0.1 R/hr at 10 ft.

Last edited: Jun 13, 2012
5. Jun 13, 2012

### sophiecentaur

How were you planning to do the energy conversion? I could only think of using a heat engine driving an alternator, which would not necessarily be very efficient. Is there some 'direct conversion' method?

6. Jun 13, 2012

### Staff: Mentor

When the gamma rays hit matter, they can produce high-energetic electrons. Maybe the efficiency can be improved if they are used directly to generate high voltage/low current, which is then converted to the needs of a lamp. In addition, the heat of the impact site can be used.
The production of 60Co is too expensive to make this practical in any way.

7. Jun 13, 2012

### sophiecentaur

There's the rub, I'm afraid. It's the actual details that would have to be right and I don't think a method exists yet. As far as I know, the only methods available (used in spacecraft) involve using a nuclear source for heat and then thermoelectric junctions - not very efficient.