- #1

jdou86

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OP warned about not providing an attempt at a solution.

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In summary: But my answer comes nothing close to the choices below.In summary, the conversation involved trying to compute the area ratio between the end of a tube and a faraway distance, with the person's answer not matching the given choices. They were asked to provide their calculations and were given advice on how to properly solve the problem. However, the person expressed frustration with the required steps and eventually solved the problem with one line of math.

- #1

jdou86

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OP warned about not providing an attempt at a solution.

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- #2

Vanadium 50

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How far did you get on this problem?

- #3

jdou86

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- #4

DrClaude

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You'll have to provide an attempt at a solution to get help.

- #5

Nugatory

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How are you using the bit about 50% being detected when the detector is placed against the face?jdou86 said:

- #6

mfb

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Please show these calculations, otherwise we can't tell what went wrong.jdou86 said:

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When the source is in the centre of the aperture, radius r, the end cap is a hemisphere of area ##2\pi r^2##. Dividing by the area of the whole sphere we get ##\frac{2\pi r^2}{4\pi r^2}=0.5##.

With the source at distance d from the centre of the aperture, the whole sphere has area##4\pi (r^2+d^2)##. The endcap's area has shrunk to not much more than ##\pi r^2##.

However, I am uncertain how the 8cm length fits in. If it is relevant then I don't understand how it can capture 50% initially.

- #9

Vanadium 50

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Your picture isn't labeled and your variables aren't defined.

You have expressions involving numbers, but no sense of what they are, where they come from, or where you are going with this.

If this is how you solve problems, when you get to Jackson or grad QM, you won't make it out the other end.

What you need to do is

- Post a
*labeled*drawing - Write down the relevant equation
- Fill in the symbolic expressions in that equation and simplify
- Now, and only now, plug in the numbers

- #10

jdou86

- 34

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Often I just know the answer, labeling are left to the engineersVanadium 50 said:

Your picture isn't labeled and your variables aren't defined.

You have expressions involving numbers, but no sense of what they are, where they come from, or where you are going with this.

If this is how you solve problems, when you get to Jackson or grad QM, you won't make it out the other end.

What you need to do is

- Post a
labeleddrawing- Write down the relevant equation
- Fill in the symbolic expressions in that equation and simplify
- Now, and only now, plug in the numbers

- #11

jdou86

- 34

- 1

And this is physics GREthere is no time for all these steps you wantedVanadium 50 said:

Your picture isn't labeled and your variables aren't defined.

You have expressions involving numbers, but no sense of what they are, where they come from, or where you are going with this.

If this is how you solve problems, when you get to Jackson or grad QM, you won't make it out the other end.

What you need to do is

- Post a
labeleddrawing- Write down the relevant equation
- Fill in the symbolic expressions in that equation and simplify
- Now, and only now, plug in the numbers

- #12

jdou86

- 34

- 1

Thanks to you I solved it with 1 line of mathharuspex said:

When the source is in the centre of the aperture, radius r, the end cap is a hemisphere of area ##2\pi r^2##. Dividing by the area of the whole sphere we get ##\frac{2\pi r^2}{4\pi r^2}=0.5##.

With the source at distance d from the centre of the aperture, the whole sphere has area##4\pi (r^2+d^2)##. The endcap's area has shrunk to not much more than ##\pi r^2##.

However, I am uncertain how the 8cm length fits in. If it is relevant then I don't understand how it can capture 50% initially.

- #13

jdou86

- 34

- 1

And fyi your response is both rude and ignorant. I don't think landau did any of the labeling crap. But discovered phase transition.Vanadium 50 said:

Your picture isn't labeled and your variables aren't defined.

You have expressions involving numbers, but no sense of what they are, where they come from, or where you are going with this.

If this is how you solve problems, when you get to Jackson or grad QM, you won't make it out the other end.

What you need to do is

- Post a
labeleddrawing- Write down the relevant equation
- Fill in the symbolic expressions in that equation and simplify
- Now, and only now, plug in the numbers

- #14

Vanadium 50

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jdou86 said:Often I just know the answer, labeling are left to the engineers

And when you don't know what the answer is? You're not going to be able to ask us when you're taking the GRE.

Anyway, It doesn't really matter if you don't want to take our advice. This will fix itself sooner or later.

- #15

jdou86

- 34

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Labeling was never the problem I think you are too arrogant to see that. Sure. So as your egoVanadium 50 said:And when you don't know what the answer is? You're not going to be able to ask us when you're taking the GRE.

Anyway, It doesn't really matter if you don't want to take our advice. This will fix itself sooner or later.

- #16

PeterDonis

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jdou86 said:fyi your response is both rude and ignorant

jdou86 said:I think you are too arrogant to see that

This attitude just got you a warning and a closed thread. Please be civil to others when you come here asking for help with a problem.

Also, for future reference, you should not post equations or solutions as images. Please use the PF LaTeX feature to post those directly:

https://www.physicsforums.com/help/latexhelp/

A gamma ray detector is a scientific instrument that is used to detect, measure, and analyze gamma rays. Gamma rays are high-energy electromagnetic radiation that is emitted from radioactive substances and nuclear reactions.

A gamma ray detector works by using a scintillator material, which converts the gamma rays into light particles. The light particles are then converted into electrical signals by a photomultiplier tube. These signals are then amplified and processed to produce a digital readout of the gamma ray energy and intensity.

The purpose of a gamma ray detector is to detect and measure gamma ray radiation in various scientific and medical applications. This includes studying the properties of radioactive materials, diagnosing and treating medical conditions, and monitoring radiation levels in nuclear facilities.

The accuracy of a gamma ray detector depends on the specific instrument and its calibration. However, most modern gamma ray detectors have a high level of accuracy and precision, with errors typically less than 1%.

In the GRE problem, the gamma ray detector is used to measure the energy of gamma rays emitted from a radioactive source. This information is then used to determine the half-life of the radioactive material and solve the problem.

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