# Question #dd2af

##### 1 Answer

#### Answer:

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

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 **per second**. Keep this in mind.

Now, the next thing to do is convert the kinetic energy of a *single* alpha particle from **megaelectronvolts**, *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

#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 **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

*half-life*,

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 **undecayed** after **one second**, and

# ( 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 **per second**.

You can thus determine how many nuclei must be present in the initial sample so that a total of

#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")#