Modern Physics Extended Response Question

[newyork]

Homework Statement

A polonium nucleus of atomic number 84 and mass number 210 decays to a nucleus of lead by the emission of an alpha particle of mass 4.0026 atomic mass units and kinetic energy 5.5 MeV. (1 atomic mass unit=931.5 MeV/c^2=1.66x10^-27kg.) The atomic number of the lead nucleus is 82. The mass number of the lead nucleus is 205.9974 atomic mass units.

a) Determine the mass difference between the polonium nucleus and the lead nucleus, taking into account the kinetic energy of the alpha particle but ignoring the recoil energy of the lead nucleus.

b) Determine the speed of the alpha particle. A classical (nonrelativisitc) approximation is adequate.

Homework Equations

I think these are the equations necessary.
a) E=mc^2
b) KE= (1/2)mv^2

The Attempt at a Solution

a) I rearranged E=mc^2 so m=E/c^2
5.5 MeV/ (3x10^8 m/s)= 6.11x10^-17 MeV/ (m/s)^2
Next, I converted to kg.
(5.5MeV/c^2)(1.66x10^-27 kg/931.5 MeV/c^2)= 1.78x 10^-30 kg

b) I rearranged KE=(1/2)mv^2 for v^2=2KE/m
KE=931.5 MeV, m=4.0026 amu
Then I took the square root and got v=21.57 m/s.

I know what I did is completely wrong, but it was the best I could figure out. Help would be greatly appreciated.

 

[dynamicsolo]

a) I rearranged E=mc^2 so m=E/c^2
5.5 MeV/ (3x10^8 m/s)= 6.11x10^-17 MeV/ (m/s)^2
Next, I converted to kg.
(5.5MeV/c^2)(1.66x10^-27 kg/931.5 MeV/c^2)= 1.78x 10^-30 kg

b) I rearranged KE=(1/2)mv^2 for v^2=2KE/m
KE=931.5 MeV, m=4.0026 amu
Then I took the square root and got v=21.57 m/s.

Part A will take a little longer to sort out. I can tell you right away on Part B that you've used the wrong value for the KE, according to the problem. The KE of the alpha particle is given as 5.5 MeV = 5.5·10^6 eV . One electron-volt (eV) is equal to 1.60·10^-12 J ; this conversion will make your velocity calculation easier.

 

[newyork]

so using the new KE (I mean it converted to ev), I got v=1.66x10^3. Did I use the right mass, or should I have used the mass I got in part a?

 

[dynamicsolo]

For Part A, it will be simpler to deal with if you work with mass-energy entirely in MeV first, then express the mass difference as MeV/(c^2) or convert it to kg., if you like.

You want to set up a conservation of mass-energy equation for this reaction:

mass of polonium-210 nucleus -> mass of lead-2?? nucleus + mass of alpha particle + alpha particle's KE (+ negligible lead nucleus recoil KE).

It's the difference (mass Po-210 - mass Pb-2??) that you're asked to find. [And the number of amu in the lead nucleus is part of what you have to figure out.]

 

[dynamicsolo]

so using the new KE (I mean it converted to ev), I got v=1.66x10^3. Did I use the right mass, or should I have used the mass I got in part a?

Please show your calculation on this part -- this speed looks too low. What units is in it?

 

[newyork]

Before you said, "One electron-volt (eV) is equal to 1.60·10^-12 J" but I thought 1 eV=1.6x10^-19 J. So, I did (5.5x10^6 eV)(1.6x10^-19J)= 8.8x10^-13 J

Then, I converted the mass of the alpha particle (4.0026 amu) to kg and got 6.62x10^-27 kg.

Finally, I plugged in the bold numbers into KE=(1/2)mv^2 and got v=16583123.95 m/s

Going back to part a, you said "It's the difference (mass Po-210 - mass Pb-2??) that you're asked to find. [And the number of amu in the lead nucleus is part of what you have to figure out.]"

The question said that the mass number of the lead nucleus is 205.9974 atomic mass units. So can't I just do 210 amu- 205.9974 amu=4.0026 amu (or 6.64x10^-27 kg)?

 

[dynamicsolo]

Before you said, "One electron-volt (eV) is equal to 1.60·10^-12 J" but I thought 1 eV=1.6x10^-19 J. So, I did (5.5x10^6 eV)(1.6x10^-19J)= 8.8x10^-13 J

Sorry, my typo: one electron-volt (eV) is 1.60·10^-12 erg; that's my old pre-SI units sneaking in when I don't proofread enough...

Finally, I plugged in the bold numbers into KE=(1/2)mv^2 and got v=16583123.95 m/s

... or about 16,600 km/sec (I get a little lower, so I think we rounded off in different places). That looks a bit more reasonable.

The question said that the mass number of the lead nucleus is 205.9974 atomic mass units. So can't I just do 210 amu- 205.9974 amu=4.0026 amu (or 6.64x10^-27 kg)?

I missed the last sentence, but I'm also a bit surprised they stated that. The difference can't be exactly 4.0026 amu, because the 5.5 MeV for the alpha particle's kinetic energy has to come from somewhere; that "somewhere" is the mass-energy of the Po-210 nucleus. So you should use the conservation equation above to figure out the mass difference, which you could just express in amu.

 

[newyork]

Thanks for the help on part b. I understand it now.

For part a: "mass of polonium-210 nucleus -> mass of lead-2?? nucleus + mass of alpha particle + alpha particle's KE (+ negligible lead nucleus recoil KE)."

This convservation equation confuses me because of the different units involved. So would the mass of lead and the alpha particle be in MeV/c^2? But then can you add this unit with 5.5 MeV?

 

[dynamicsolo]

For part a: "mass of polonium-210 nucleus -> mass of lead-2?? nucleus + mass of alpha particle + alpha particle's KE (+ negligible lead nucleus recoil KE)."

This conservation equation confuses me because of the different units involved. So would the mass of lead and the alpha particle be in MeV/c^2? But then can you add this unit with 5.5 MeV?

When we talk about the mass-energy of the particles as mass, the factor c^2 has to be introduced as a conversion factor under "mass-energy equivalence". You'll notice that the atomic mass unit is given as 931.5 MeV/c^2 = 1.66x10^-27kg . (It's a bit like what gets written for the metric-to-English conversion 1 kg. = 2.205 lb. One kilo is, of course, not equal to any number of pounds because it's a unit of mass and the pound is a unit of force. But there is an understanding that 1 kilogram has the weight-equivalent of 9.81 N in "Earth gravity" and that 9.81 N = 2.205 lb.)

What you are really adding in the conservation equation (and perhaps I should have written it that way) is

mass-energy of polonium-210 nucleus -> mass-energy of lead-206 nucleus + mass-energy of alpha particle + alpha particle's KE (+ negligible lead nucleus recoil KE) .

You would now (implicity) multiply the 210 amu of the polonium nucleus by c^2 to get an energy in MeV and use (210 · 931.5 MeV/c^2 · c^2), and so forth. For our problem, since we want the mass difference of the polonium and lead nuclei, we can just look at the sum

mass-energy of alpha particle + alpha particle's KE ,

make the appropriate adjustment for the alpha, add the energies, and convert the result to a mass.

 

[newyork]

What you are really adding in the conservation equation (and perhaps I should have written it that way) is

mass-energy of polonium-210 nucleus -> mass-energy of lead-206 nucleus + mass-energy of alpha particle + alpha particle's KE (+ negligible lead nucleus recoil KE) .

You would now (implicity) multiply the 210 amu of the polonium nucleus by c^2 to get an energy in MeV and use (210 · 931.5 MeV/c^2 · c^2), and so forth. For our problem, since we want the mass difference of the polonium and lead nuclei, we can just look at the sum

mass-energy of alpha particle + alpha particle's KE ,

make the appropriate adjustment for the alpha, add the energies, and convert the result to a mass.

Thank you so much for the help. That equation made it a lot clearer. I'm going to show my teacher the work tomorrow and see if she approves.

 

[dynamicsolo]

I perhaps should just mention a point about the mass-differences. If you look at a reference for the masses of nuclei, you find that their masses in amu are less than you would expect if you added up the masses of the protons and neutrons present. The reason seems to be that the binding field (the force between the constituent quarks of the protons and neutrons) contributes a negative energy (the so-called "binding energy") to the nucleus. In terms of mass-energy equivalence, this appears as a negative mass contribution to the measured mass of the nucleus. (This is also why the atomic mass unit -- defined as 1/12 the mass of a carbon-12 nucleus -- is smaller than the mass of a proton or a neutron.)

In the case of the nuclei discussed in your problem, the "mass" of the lead-206 nucleus is slightly lower than you would think it should be from simply removing the alpha particle. The nucleons in the lead-206 nucleus are a bit more tightly bound than those of the polonium-210 nucleus (if this weren't so, the polonium wouldn't be unstable and thus "radioactive") and the excess energy released from the reconfiguration of nucleons goes into the kinetic energy of the alpha particle (and, to a lesser extent, into the recoil energy of the lead nucleus).