Modern Physics Study Questions

Bring on the tough stuff.

1. Why do stars build an Iron core over time?

1. Answer: Fusion of Hydrogen into Helium, then Helium into heavier elements stops at Iron, because Iron has the most stable, most tightly bound nucleus. Incidentally, building an Iron core eventually kills the star because it means it's running out of fuel to burn. Burning Hydrogen gives it the biggest energy gains, as shown by the graph below.

2. Hypothesize why the n = 1 level is so much deeper of an energy well than n = 2, with a larger energy gap, than any other two energy levels.

2. Answer: While there's plenty of quantum physics to the full answer to this question, the electromagnetic force has a lot to do with it. Lower energy levels are closer to the nucleus, so the force between the negative electron and a positive nucleus is larger. We might also hypothesize that if the radius for n = 2 and n = 3 orbitals add a linear increase, then we'd expect that the difference in energy levels between levels would fall off as } as Coulomb's Law indicates.

3. A green photon with a wavelength of 550 nm is emitted from an atom when an electron jumps from an n = 5 level to an n = 2 level. What's the energy difference between the two levels in eV?

3. Answer: All we need to know is that E = hf, because that's the same as the energy difference in question. Phew! We find the frequency of the light using c = fλ, so f = 5.45 × 10¹⁴ Hz. Next, ΔE = (6.626 × 10^-34 J∙s)(5.45 × 10¹⁴ Hz) = 3.61 × 10^-19 J, which after converting Joules into electron volts by dividing by e, is 2.26 eV.

4. A Germanium nucleus exchanges two of its neutrons for protons while decaying into Selenium:. This is called double-beta decay.

Formulate a hypothesis as to why some elements might undergo double-beta decay as opposed to single-beta decay.

4. Answer: Everything we've ever studied in physics teaches that nature always takes the easiest path to get to where it needs to be. Included in this is Newton's Law of Inertia¹⁴ and why electricians put in extra resistors temporarily while setting up circuits¹⁵. In the context of nuclear physics, atoms and nuclei will do whatever they can until they're back to their comfortable ground state. They will not stop, they will never give up and they will not quit. They are tenacious little buggers who hate carrying extra energy, because, well, it tires them too much.

Single beta-decay occurs when a nucleus has an excess of neutrons and flips them into protons. The nucleus is a lot more stable and a lot happier too. Although some effort was required, in the end, the nucleus is in its lowest possible state. It's better off putting in that little extra effort so that in the long run, it's happier.

Our instincts tell us it'd require more energy to turn two neutrons into two protons rather than one. But we know that can't happen because nuclei are waaay too lazy. So if double-beta decay occurs, it means single-beta decay can't happen, because it would put a nucleus into an even more excited state.

5. Write an equation for a hypothetical double-decay that does not release neutrinos. This is called neutrino-less double-beta decay. If this decay was to be observed, what laws would be violated?

5. Answer: Let's stick to Germanium while we're at it, though other nuclei can also undergo this rare double-beta decay. So, starting with , we can omit the two anti-neutrinos and write .

What laws are broken? Since the first equation is perfectly balanced, we are tempted to write angular momentum, energy and mass as starters. However, this type of decay could exist if the two neutrinos annihilated each other. The reasons go beyond the scope of this module and some particle physicists disagree on this theory.

Baryon charge is conserved with an equal number of protons and neutrons. So is electrical charge, since we have two extra protons with two electrons as products. What's missing? You got it– leptonic charge is no longer conserved, because of those two electrons on one side of the equation and without the two anti-neutrinos.

Seems strange, we know.

6. As we look at a periodic table, we notice that each stable element has an equal number of neutrons and protons up until Calcium, . After that, there is a clear departure as the number of neutrons per proton increases. Using our knowledge of fundamental forces and nuclei size, explain this nucleon number asymmetry. In other words, why are there so much more neutrons than protons as we go down the periodic table?

6. Answer: Remember that the radius of a nucleus is directly proportional to A, the mass number with our R = R_oA^1/3 equation. This means a nucleus will need to be larger to host more nucleons. Sure, that makes sense, but that doesn't explain why there are more neutrons than protons.

Protons and neutrons are both baryons. Protons have a positive charge and neutrons are neutral. Both protons and neutrons follow the guidelines of the strong nuclear force. So while protons are tugged towards their electrical counterparts the electrons and repulsed by their fellow protons, they're also tugged back to the nucleus through the strong force. An atom gets bigger and collects more electrons; the only thing a nucleus can do to save its life is to invite more neutrons into its cozy nest.

7. The proton-proton chain that fuels the Sun fuses Hydrogen atoms into Helium, . Calculate the binding energy of this Helium atom knowing that its nuclear mass is 4.001505 u.

7. Answer: The binding energy E_b of a nucleus is given by E_N + E_b = Zm_pc² + (A − Z)m_nc², where (A − Z) is the number of neutrons, E_N is the energy of the nucleus, Z is the number of protons and m_p is the proton mass.

Let's ignore c² for now to make things simpler. We have the mass of 2 protons, or 2m_p = (2 × 1.007825 u) = 2.014552 u. We have the mass of 2 neutrons, or 2m_n = (2 × 1.008665 u) = 2.017330 u. We disregard the electrons as they don't inhabit the nucleus nor influence its binding energy.

We add the two to get the total: 2.014552 u + 2.017330 u = 4.031882 u. The difference between E_N and E_b is then (4.031882 u − 4.001505 u) = 0.030377 u. Knowing that , we find a binding energy of E_b = 28.3 MeV.

8. Astrophysicists have plotted the average binding energy per nucleon versus a nucleus' mass number A, as seen in Question 1 and elsewhere in this unit. How does this compare to our 28.3 MeV helium result?

8. Answer: There are 4 nucleons in helium. Dividing 28.3 MeV by 4 is 7.07 MeV per nucleon. It looks like we did a great job, since that's what the graph tells us too.

9. The proton-proton chain that fuels the Sun fuses Hydrogen atoms into Helium. Although the chain is extensive, in the end, 4 protons are needed to form the final product of a Helium atom, . The reaction can be summarized as:

Keeping in mind matter annihilation and assuming neutrinos have no mass worth mentioning, calculate the energy released from this reaction.

9. Answer: So, let's not panic. That can be the first step of the chain.

We have 4 protons and 2 electrons on the left side. Let's start with that. The mass of 4 protons is simply 4m_p = (4 × 1.007825 u) = 4.029104 u. The mass of 2 electrons is 2m_e = (2 × 5.49 × 10^-4 u) = 0.001095 u.

That's a total of 4.030199 u for the mass of the reactants.Now let's move on to the right side of the reaction.

Photons have no mass, and neutrinos are considered to be massless for these calculations. The positrons have the same mass as the electrons on the left side, so we can cross off both β terms. That means we only have one Helium atom to worry about on the right side. See? In a matter of seconds, we just simplified the problem a great deal.

We stated earlier on the nuclear mass of a helium atom is 4.001505 u. That's the mass we need to use to calculate the energy released. We take the mass difference of Helium from the four protons, (4.030199 u − 4.001505 u) = 0.028694 u. Since , the energy released is equal to 26.7 MeV.

<strong>10.</strong> Massive stars die by collapsing upon themselves and turning into a supernova. Neutron stars sometimes form after supernova explosions and are essentially made up of neutrons with the density of a nucleus because the electrons and protons combine to form neutrons. Calculate the radius of a neutron star if its mass is equal to the Sun's, about2 × 10³⁰ kg.

10. Answer: First we need to calculate the density ρ of nuclear matter, using our old friendly equation . Let's approximate the mass of a nucleus to equal the proton's mass, which is m_p = 1.67 × 10^-27 kg. A nucleus with a number A of protons will weigh A × 1.67 × 10^-27 kg.

Now, what is the size R of a nuclear radius again? Roughly R = R_oA^1/3, where R_o = 1.2 fm, or 1.2 × 10^-15 m and A is the mass number of the nucleus. Since the volume of a sphere V is given by , we then write out the density of nuclear matter as .

Since the Sun's mass is 2 × 10³⁰ kg, then the volume of a neutron star with that same mass would just equal .

One more step and we're done, we promise. We need to extract the radius out of the volume with . Rearranging everything, we get . That's DENSE.