The energy released by αdecay in a 50.0 -g sample of 94^299 Pu is to be used to heat 4.75 kg of

The energy released by αdecay in a 50.0 -g sample of 94^299...

Key Concepts

Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. In this context, it refers to the spontaneous transformation of a nucleus into a different state or element by expelling particles, which results in a release of energy. This process is fundamentally probabilistic and follows decay laws that describe the rate at which a sample undergoes decay.

Alpha Decay

Alpha decay is a type of radioactive decay in which an unstable nucleus emits an alpha particle, which consists of two protons and two neutrons. This emission results in a reduction of both the atomic number and mass number of the original nucleus, and the energy released in this process is typically large due to the nuclear binding energy differences between the initial and final states.

Energy Conversion from Nuclear Reactions

When a nucleus undergoes decay, the energy released is initially in the form of kinetic energy of the emitted particles. This energy can be converted into thermal energy if absorbed by a material. Converting this nuclear energy into heat requires understanding the total energy released from all individual decay events over a given time period.

Specific Heat Capacity

Specific heat capacity is the amount of energy required to raise the temperature of a unit mass of a substance by one degree Celsius. In the context of heating water, the specific heat capacity of water is used to determine how much the temperature of the water will increase when it absorbs a given amount of energy. This concept is critical in relating the energy input from radioactive decay to the resulting temperature change of the water.

Transcript

00:01 All right, question 88 asks, the energy released from an alpha decay in a 15 gram sample of plutonium 239 is to be used to heat 4 .75 kilograms of water.

00:12 Assuming that all the energy released by the radioactive decay goes into heating the water, find out how much the temperature of the water increases in one hour.

00:21 So it sounds like a simple, straightforward answer, which, i mean, question at least, i mean, it is.

00:27 I'll take a little bit of work here, but we're going through this.

00:31 As for the change in temperature, so if you remember from our thermodynamics, we know that the relationship between energy and change in temperature is simply the energy inserted into the system equals a mass times the specific heat of water times the change of temperature.

00:49 So we're given mass, we're given cv, of course, our specific heat of water.

00:53 We're solving for t, so all we need to find is the amount of energy released in a 50 gram sample of plutonium.

01:00 And the rest should be easy, right? right.

01:04 So first we need to figure out how much energy we need.

01:07 We need to really figure out what reaction we're dealing with.

01:10 So like you said, it's plutonium 239, and it's stated that it's an alpha particle that gets released.

01:21 So you know alpha, of course, is just the helium atom.

01:26 And if we balance this reaction, we find that we have uranium as our final product, uranium, to be precise.

01:36 Which is hopefully by now obvious that it's 239 minus 4, we get 235.

01:41 You get uranium.

01:43 So we need to look at first to see how much energy is released in each of these decay reactions, decays in general.

01:52 So you know that our energy is just the change in mass times c squared, it relates to the absolute value.

02:01 So our first step is to figure this part out and each of these masses in our appendix f here we find there is a mass of helium the mass of uranium 235.

02:20 We'll subtract off the mass of plutonium.

02:26 We don't need to take into account the mass of the electrons because it will be balanced on each sides of these equations.

02:31 This is units of u to convert to energy.

02:34 These are a factor 931 .5, mev per c squared view.

02:41 And finally multiply that by c squared.

02:43 Can't solve our units and its equation already.

02:48 You find that the amount of energy per decay is 5 .2.

02:52 Equals 5 .24 m .ev.

02:56 Remember that's for a single decay.

02:57 So i'll say m .e .v.

02:58 Per decay.

03:03 So that's a single decay.

03:04 Now we need to figure out how many of these decays will occur for a 50 gram sample of plutonium.

03:15 So that means that the number of decays, i'll just call them deck here for short, that equals our number of original samples or original atoms minus some amount.

03:27 To get down to zero.

03:36 So some original amount down to its final amount.

03:41 Well, that's easy enough to know that this n is just the same as our decay equation, our original value times e to the negative lambda t.

03:52 And we can simplify this one more time, pull out n0, 1 minus a negative lambda t.

04:02 So because we're not waiting for the whole mass or whole amount of plutonium to decay, we're only doing this for one hour.

04:10 So that's why we need to look at this time frame of this difference in time for amounts, not just using all 50 grams because they, i mean, it may happen that time.

04:20 I suspect that it won't, but in the scenario that it doesn't, we need to take that into account as well.

04:26 So now we're left with, okay, well, what is our initial amount? so we're given that we want the initial amount here for our decay equation to work.

04:36 So this amount, we can think of it as our given mass over our, molar mass, multiplied by avagadro's number at a.

04:47 Right, and just to double check that, the units of this i'll put in blue.

04:52 Well, our mass isn't grams, of course, or it can be grams, anything really.

04:57 Our molar mass is number of grams per mole.

05:02 We're multiplying by avogadro's number, which is number of atoms, or in this case, it could be number of decays, numbers, number of whatever per mole.

05:14 So we're looking at our units, we have moles will cancel, grams will cancel, and we're left with atoms.

05:19 So this is good.

05:20 Just check in our brains that this is the right equation we want to use.

05:24 So for this, for our plutonium, it's 50 grams, sample.

05:31 Our molar mass, which we should previously, is 239.

05:40 0 .052158.

05:49 That's supposed to be an 8.

05:53 58 gram per mole.

05:56 And multiply that by avagato's number, which is, of course, known 6 .022 times 10 to the 23.

06:08 Atoms decays per mole, whichever you like again, mole.

06:14 We know our units cancel out beautifully, so we don't need to worry about that this time, but we do know that the number of atoms present in the sample is 1 .26 times 10 to the 23 atoms, which also be, of course, the number of decays that will occur as well.

06:37 Or initial, sorry, our initial number of events, not number of decays.

06:41 It could be, but it's not, because we'll know that.

06:44 We'll touch a long half -life and won't be the case.

06:46 Anyway, so next part of this equation that we don't have is lambda.

06:54 So we can solve for that as well...

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