Question dd2af

Nov 27, 2015

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 $\text{1400 W}$ 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

${\text{1 W" = "1 J s}}^{- 1}$

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

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

$\text{1 MeV" = 1.60217662 * 10^(-13)"J}$

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 $\text{1400 J}$ of energy?

$\textcolor{b l u e}{\alpha} \cdot \text{energy per alpha" = "total energy}$

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 $1.5702 \cdot {10}^{15}$ 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

$\textcolor{b l u e}{A = {A}_{0} \cdot \frac{1}{2} ^ n} \text{ }$, where

$n = \frac{t}{t} _ \text{1/2}$ - the ratio between the amount of time that passed, $t$, and the half-life, ${t}_{\text{1/2}}$, of the substance

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

$\text{1 year " = " 31,556,926 s}$

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

$A = {A}_{0} \cdot \frac{1}{2} ^ \left(\frac{1}{\text{2.767,542,410.2}}\right) = {A}_{0} \cdot 0.9999999997495441$

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 $\textcolor{b l u e}{\alpha}$ 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

$m = \textcolor{g r e e n}{\text{2400 g}}$