Question

Here we have a Radioactive Thermoelectric Generator (RTG) that is powered by Plutonium 238 (²³⁸Pu). The RTG produces 1400 Watts. ²³⁸Pu decays by alpha-particle emission. The alpha-particle has 5.593 MeV of kinetic energy. The half life (t_1/2) of ²³⁸Pu is 87.7 years.

a) Write the nuclear reaction for ²³⁸Pu decay

b) Assume all of the kinetic energy is converted to heat. How much ²³⁸Pu (in grams) must be present to produce 1400 Watts of power?

Answer 1

a)

²³⁸Pu → ²³⁴U + alpha particle.

b)

wes need to use the kinetic energy per alpha particle to determine how many alpha particles we would need in order to get 1400 W of power.

Once we know that,we will use plutionium-238's half-life to determine how many alpha particles are emitted per second for a given mass of the isotope.

Finally, we will use the number of emitted alpha particles to determine how many nuclei of plutonium-238 you'd need to have in the sample.

So, it's important to realize here that Watts are equivalent to Joules per second

1 W=1 J s−1

This means that in order to generate a power of 1400 W, you need to be able to produce 1440 J per second. Keep this in mind.

Now, the next thing to do is convert the kinetic energy of a single alpha particle from megaelectronvolts, MeV, to Joules by using the conversion factor

1 MeV=1.60217662⋅10−13J

This will give

5.593MeV⋅1.60217662⋅10−13J1MeV=8.961⋅10−13J

So, number of alpha particles must be emitted per second in order to produce a total of 1400 J of energy:

α⋅energy per alpha=total energy

α=1400J8.916⋅10−13J=1.5702⋅1015alpha particles

This means that the mass of plutonium-238 must emit 1.5702⋅1015 alpha particles per second in order to allow for that much energy to be produced.

Now it's time to use the nuclear half-life equation to determine how many nuclei will decay in one second

A=A0⋅12n , where

n=t1/2 - the ratio between the amount of time that passed, t, and the half-life, t1/2, of the substance

Convert the half-life of plutonium-238 from years to seconds by using the conversion factor

1 year = 31,556,926 s

In this case, you would have

87.7years⋅31,556,926 s

1year=2.767,542,410.2 s

This means that we have

A=A0⋅1212.767,542,410.2=A0⋅0.9999999997495441

Now, if A represents the amount of an initial sample that remains undecayed afterone second, and A0 represents the initial mass of the sample, you can say that

A0⋅0.9999999997495441A0×100=99.99999997495441%

of the initial sample remains undecayed after one second. This of course means that only

100%−99.99999997495441%=0.000000025046%

of the initial sample will decay in one second. This means that out of 100 nuclei of plutonium-238, only 0.000000025046 nuclei will decay per second.

You can thus determine how many nuclei must be present in the initial sample so that a total of α particles are emitted in one second by

1.5702⋅1015alpha particles⋅100 nuclei0.000000025046alpha particles=6.0177⋅1024nuclei

Helpful 0

Confusing 0

Now all you have to do is use Avogadro's number to determine how many moles of plutonium-238 would contain this many nuclei, and the isotope's molar mass to determine how many grams.

6.0177⋅1024nuclei⋅1 mole238Pu6.022⋅1023nuclei=9.999286 moles238Pu

Finally, this is equivalent to

9.999286moles⋅238.05 g1mole=2378.8 g

Rounded to two sig figs, the number of sig figs you have for the power of the RTG, the answer will be

m=2400 g