Question

Question #dd2af

Answer 1

Now this is an interesting problem!

Explanation:

The idea here is that you need to use the kinetic energy per alpha particle to determine how many alpha particles you would need in order to get `"1400 W"` of power.

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

Finally, 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^(-13)"J"`

This will get you

`5.593 color(red)(cancel(color(black)("MeV"))) * (1.60217662 * 10^(-13)"J")/(1color(red)(cancel(color(black)("MeV")))) = 8.961 * 10^(-13)"J"`

So, how many alpha particles must be emitted per second in order to produce a total of `"1400 J"` of energy?

`color(blue)(alpha) * "energy per alpha" = "total energy"`

`color(blue)(alpha) = (1400color(red)(cancel(color(black)("J"))))/(8.916 * 10^(-13)color(red)(cancel(color(black)("J")))) = 1.5702 * 10^(15)"alpha particles"`

This means that the mass of plutonium-238 must emit `1.5702 * 10^(15)` alpha particles per second in order to allow for that much energy to be produced.

SIDE NOTE The following calculations will look horrendous - don't be intimidated! Use a calculator and keep track of the decimals you get in every step!

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

`color(blue)(A = A_0 * 1/2^n)" "`, where

`n = t/t_"1/2"` - the ratio between the amount of time that passed, `t`, and the half-life, `t_"1/2"`, 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 your case, you would have

`87.7 color(red)(cancel(color(black)("years"))) * "31,556,926 s"/(1color(red)(cancel(color(black)("year")))) = "2.767,542,410.2 s"`

This means that you have

`A = A_0 * 1/2^(1/"2.767,542,410.2") = A_0 * 0.9999999997495441`

Now, if `A` represents the amount of an initial sample that remains undecayed after one second, and `A_0` represents the initial mass of the sample, you can say that

`( color(red)(cancel(color(black)(A_0))) * 0.9999999997495441)/color(red)(cancel(color(black)(A_0))) xx 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 `color(blue)(alpha)` particles are emitted in one second by

`1.5702 * 10^(15)color(red)(cancel(color(black)("alpha particles"))) * "100 nuclei"/(0.000000025046color(red)(cancel(color(black)("alpha particles")))) = 6.0177 * 10^(24)"nuclei"`

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 * 10^(24)color(red)(cancel(color(black)("nuclei"))) * ("1 mole"""^238"Pu")/(6.022 * 10^(23)color(red)(cancel(color(black)("nuclei")))) = "9.999286 moles"""^238"Pu"`

Finally, this is equivalent to

`9.999286color(red)(cancel(color(black)("moles"))) * "238.05 g"/(1color(red)(cancel(color(black)("mole")))) = "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 = color(green)("2400 g")`