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Tutor profile: Philip B.

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Philip B.
Senior In Nuclear Engineering At The University of Michigan with a specialization in detector physics
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Questions

Subject: Nuclear Physics

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Question:

Explain why neutrinos were originally deemed necessary and thus confirmed to exist before their first detection.

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Philip B.
Answer:

Neutrinos are most commonly seen in beta decays (both beta plus and beta minus) where $$\beta^-$$:$$ n \rightarrow p+e^-+\bar{\nu_e}$$ and $$\beta^+$$:$$p \rightarrow n + e^+ +\nu_e$$. These two interactions are what sparked the need for neutrinos. The initial discovery was due to the necessity of a third particle to conserve momentum in the system, as the energies of the detected electrons and positrons in these interactions occupied a range, which in order to conserve the 0 momentum state of the parent nucleus, there would have been finite values for each interaction, as there would be only one possible energy distribution. However, if momentum could be transferred to another particle, we could observe this range, as momentum could be divided between the three in many different ways within individual geometries (of which there could be multiple) and still be able to conserve momentum.

Subject: Nuclear Engineering

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Question:

For a scintillation based detector system comprising of multiple detectors observing a spontaneous fission source, what are some potential issues that can affect the perceived multiplicity of the system? Assume that the detector is 100% efficient in detecting all incident particles.

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Philip B.
Answer:

If the detector is perfectly efficient, which is not a reasonable assumption to make as most scintillation detectors are only on the order of 20% efficient, the only factors that could alter the detected multiplicity would be incorrect detection, or lack of detection of particles emitted in fission. The second factor is easier to explain so we will start with that. As the system does not likely have a 4 pi solid angle; that is it does not perfectly encapsulate the fission source, some neutrons or photons will escape our system without interacting with a detector due to the geometry at which they are emitted, as fission events are random, and can emit particles in any direction. The first issue mentioned can be broken into two separate cases; cross talk, and room return. Cross Talk occurs when a neutron or photon is incident on one detector, is detected, and scatters into another detector and is detected by that detector as well. This is most common for detectors directly adjacent in an array, as the geometry best allows for cross talk between adjacent detectors. This is problematic as the system will see these events recorded as two separate particles emitted as part of the same event, when in reality it is only one. Room return occurs when a particle is emitted from the source, and may or may not be detected, exits the detector system, and then is scattered off of the surroundings in the room, and then re-enters the detector system, and is detected. This can affect multiplicity if the particle is detected within the timing window of other particles in the system. The system will perceive it as being part of the event that produced the other particles when in reality it is not, thus affecting the multiplicity.

Subject: Advanced Physics (Special Relativity)

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Question:

Muons are created in the upper atmosphere through the interaction of cosmic rays with the atmosphere. Muons have a mean lifetime of 2.2$$\mu$$s ($$2.2\times 10^{-6}$$s) and a mean velocity of 0.995c (0.995 times the speed of light). Based on these values, the distance a muon would travel before decaying on average would be 656.7m. We know in actuality they travel several thousand meters to reach the earth's surface as we are able to detect them. Explain how this is possible using special relativity.

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Philip B.
Answer:

The key to answering this question is by considering the two frames of reference that are in play; the frame of reference of the muon itself (observer 1), and the frame of reference of an observer stationary with respect to the surface of the earth (observer 2). We know proper time is described of being motionless with respect to the object to whom the action is occurring, and proper length is the length measured by an observer not moving relative to the axis of motion. In this case proper time is measured by observer 1 and proper length is measured by observer 2. We can approach answering this question using either frame of reference, as the muon travels the same speed in both frames, but both will be provided below: Explanation 1: Time dilation considering the motion of the muon, the time observed by observer 2 on earth will be the dilated time; $$t' = \gamma t_0$$ where $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$\gamma$$ for this problem $$= \frac{1}{\sqrt{1-\frac{(0.995c)^2}{c^2}}} = 10.01$$ so our mean lifetime observed by observer 2 is 10 times greater than that of observer 1, while traveling at the same velocity, meaning it is able to travel 10 times further, as it exists for ten times longer. Explanation 2: Length Contraction Considering the motion of the muon, observer 2 will observe proper length $$L' = \frac{L_0}{\gamma}$$ so in this case, the length that the muon perceives that it travels is 10 times less than that observed by an observer on earth, meaning that while the muon percieves itself traveling only aproximately 650m, an observer on earth observes it traveling 6.5km Using either explanation, we can see that the muon has the ability to travel from earth's atmosphere to a height at which it can be observed in a detector system.

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