I was studying on partial derivatives, chain rule etc uptill now. Deadline of the homework is tomorrow night. I guess I will not be able to complete the homework. If someone would give me the solution I really appreciate it. Else I will quit the question. I know that here is not a question...
This is a piece of a spectrometer question and this region above is the velocity selector region. The question says "What should be the power of the voltage supply that we will use in velocity selector region?".
I think it is not a problem to just take this area about the question. I assumed...
I didn't ask the question for anything about the kinetic energy of the particle. You misunderstood me . I just wondered that what is the voltage supply's power.
I know only these relevant equations for power. But here there is a different situation, we have a parallel plates which we don't have to know I current. How should I think of about this? I wonder if it can be thought of a capasitor or something?
Assume that also we know E, B and velocity .
In velocity selector region, the question says "determine" but I did it like this below. I chose the voltage necessarily. Is there another way to determine these values?
I'll be hanged if I understand! Pardon but I'm just not be able to do these calculations. I'll glance at chain rule of partial derivatives. I guess we couldn't make progress.
I was talking about the same thing at #31. And because of we can't use these as m2-m1 in ##
(r_2 - r_1) = r_2 (1 - \sqrt{\frac{m_1}{m_2}})## I calculated these m2 and m1 one by one. So in this way, we don't have to use actual mneutron I think. What do you think?
Oh, I thought I fixed it when I see it first time. So, actually 1AMU = 1.66X10-27 and m1 = 205.974*1.66x10-27 = 3.4191684×10-25 = 3.42x10-25. But I can't edit the post #27 now.
It may says the same result now I guess.
I'm going to take care of it. Thanks for notice.
edit: I think only latest...
Was the difference between m1 and m2 supposed to be 1.67x10-27? I wonder if I made the wrong calculation.
m2 - m1 = 0.02x10-25 = 2x10-27. Does a 0.33 slip cause a problem? I hope does not :)
I can't give an example because of it doesn't make sense to me.
I guess I found ## \partial Ψ##/ ## \partial \ y##. But I don't get why we are doing these calculations somehow. The question asks for ## \partial \ u##/ ## \partial \ x## etc.
I chose second one and found the answer like this. Idk whether it is incorrect. So I'll try for the first one for the same logic, I'll see if it will get the same answer.
Edit: from first one, because of there is no natural logarithmic situation, we can't get ln2 or something
I glanced the link and knew the drill. But I didn't find u -the scalar potential- of ## \ F = (\frac{xy+1}{y},\frac{2y-x}{y^2})##.(so λ =1) I'll really appreciate it if someone calculate the u. I think it is not such difficult to someone who knows the progress. The problem is I am learning...
It's a great expression with the animation. Those glittered when I read :
qE = qvB
E = vB
v = E/B
So while I choose the E field and B field, I will pay attention to protect the ratio between E and B. For Pb207 (which I've introduced as 2 in #27), with v2 = 0.77x106m/s the ratio will 0.77x106...
In spectrometer region:
because of (r2 - r1 )= r2(1−√m1/m2) --> 0.01 = r2(0.003) --> r2 =3.33m, r1 = 3.32 m.(m1 = 3.42*10-27x205.974 = 3.42*10-25, m2 = 3.42*10-27x206.975 = 3.44*10-25)
2 for Pb207, 1 for Pb206 so I chose first one(2 (Pb207))
So FB = q(vxB) = qvB and circular force = FC =...
I wonder if it is expected to create a mass spectrometer which have all things from us and if ;
whole things in the spectrometer will be defined from us. but relative to what?
After a short break, I'm here again. I would like to tell the process;
I selected Pb207-Pb206 isotopes combination and did the calculation as selecting the isotopes masses(amu). So we know the difference in radius between two isotopes r=0.01m, B=0.5T, Qe = 1.60*10-19 , and r= (m.v)/(q.B)...
it again escaped me. So now can I take the magnetic field as 1.0 Tesla or maybe less?
I don't actually get what it says. How far must I choose?
edit: The magnetic field of a typical refrigerator magnet is;
5 × 10−3 T (5 mT) – the strength of a typical refrigerator magnet
Which type of magnet...
I have a trouble with square shape symmetry. How can I come through of it? Or maybe I don't have to show calculations, just draw a figure which can explain the logic.
Yeah it is obvious but I must attach an explain and show the calculations. Have you any offer to do it by this way? Or maybe a figure which can explain.
"In order to achieve this, choose a magnetic field with a magnitude in Tesla (maximum magnetic field you can obtain from a conventional magnet is around 2.5 T so be far away from this value) and choose the direction also."
I consider the bold statement, for instance in spectrometer or velocity...
Okay I get it now, so it must be either Pb206-Pb207 or Pb207-Pb208 right? Supposing that I chose second combination, then ;
r2 - r1 = (m2-m1)*(v2-v1)/e*B.
We know that m𝑛𝑒𝑢𝑡𝑟𝑜𝑛=1.67×10−27 𝑘𝑔 , 𝑄𝑒=1.60×10−19 𝐶𝑜𝑢𝑙𝑜𝑚𝑏𝑠 and
.
But we don't know (v2-v1) and B. If we are right thus far, what is the...
An ionizing laser is directed onto the lead block under test such that the ionized atoms (or isotopes) leaves the lead block with an initial kinetic energy of 1 eV. The direction is conically upward but sure that the cone is larger than the entrance hole to the accelerator section. The ionized...
The question is here. I thought someone with knowledge of spectrometers could easily answer the question I asked at #7, but it's OK.
1.The mass difference between two isotopes is sometimes just a neutron mass. The spectrometer should separate them very well. For such an isotope combination, the...
I read a lot of articles and texts about spectrometers. Even though the system I had is different than a classic spectrometer, I understood the logic except 1μA. I think it is a significant thing for solving the whole problem because of there is no enough information (for instance, the voltage...
Thank you for informations. I'm going to study on all of them.
I also study on it. Do you know any source about spectrometers? If not I just continue with Serway
Summary:: How to load the plates with the Lead element in the spectrometer
I have a mass spectrometer with lead element which has an electronic configuration 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 4f^14 5d^10 6s^2 6p^2. It has 2 free electron, so the ejected electrons go...
The final solution is this then:
##\frac{k\ lambda}{(y_p-a)}\ =\frac {(x_p+x)}{[(x_p+x)^2+(y_p-a)^2]^\frac{1}{2}} ## with borders from ##-a## to ##a##
I don't know where ##(y_p-a)## in denominator came from in the linked page at #2. In link ##z## = ##(y_p-a)##
now. Please tell me that is true...
##\int_{x=-a}^a\frac {(y_p-a).dx}{((y_p-a)^2+(x_p+x)^2)^\frac 32}##
##k\ lambda\ (y_p-a)\int_{a-x_p=-A}^{a+x_p=B}\frac {d(x+x_p)}{((y_p-a)^2+(x+x_p)^2)^\frac 32} =\frac {(x_p+x)}{[(x_p+x)^2+(y_p-a)^2]^\frac{1}{2}} ## with borders from ##a-x_p## to ##a+x_p##
Is this right?
Note: I cheated...
Can't we solve this by not touching on matching ##k\lambda (y_p-a)\int_{x=-a}^{x=a}\frac{dx}{((x+x_p)^2+(y_p-a)^2)^{\frac 32}}## and ##k\lambda z\int_{t=-A}^{t=B}\frac{dt}{(z^2+t^2)^{\frac 32}}##. Because I couldn't understand replacing A and B in the linked page somehow. I mean isn't there any...
I was studying on other lectures. So I wasn't here for a month. I just glanced all comments above and I decided to choose a point on the upper line this time.
##dE_y=dE \sin\theta= \ k\ y_p\lambda∫ \dfrac{ dx}{\left[(x+x_p)^2+(y_p-a)^2)\right]^{3/2}}##
##dE_y= \ k\ y_p\lambda∫ \dfrac{...
Yeah, I was actually going to ask only the surface rules but I forgot where I am and just continue. Can you transport this thread there? If you can't I will post a new thread then
I asked to confirm this solution but solvent had not apply any integration to find the electric field. So I'm not sure whether the solution is correct.
Summary:: For finding the electric field at P in the photo below, may I select a gaussian surface circular?
[Mentor Note -- thread moved to the schoolwork forums, so no Homework Template is shown]