June 2011 G485-Fields, Particles and Frontiers of Physics Watch

Pandit Bandit
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can anyone summarise the alpha scattering experiment for me please?
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Oh my Ms. Coffey
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(Original post by Pandit Bandit)
can anyone summarise the alpha scattering experiment for me please?
Originally it was thought the atom was of a plum pudding model.

The Rutherford experiment disproved this. Gieger used an Alpha emitting source encased in a lead block which would fire He nucleus's towards a very thin piece of gold foil. A curved florescent screen was placed behind which would flash when hit by the alpha particle.

Most of the times alpha particles were deflected by a slight amount however occasionally they were deflected by an angle greater then 90 degrees.

From this it was deduced that the gold atom was mostly open space and the minor deflections were caused by repulsion from a positively charged nucleus. The major deflections were due to the nucleus of the gold atom being much larger causing the alpha particle to rebound so the atoms nucleus must be tiny but have a lot of mass.
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sulexk
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#83
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(Original post by Oh my Ms. Coffey)
Originally it was thought the atom was of a plum pudding model.

The Rutherford experiment disproved this. Gieger used an Alpha emitting source encased in a lead block which would fire He nucleus's towards a very thin piece of gold foil. A curved florescent screen was placed behind which would flash when hit by the alpha particle.

Most of the times alpha particles were deflected by a slight amount however occasionally they were deflected by an angle greater then 90 degrees.

From this it was deduced that the gold atom was mostly open space and the minor deflections were caused by repulsion from a positively charged nucleus. The major deflections were due to the nucleus of the gold atom being much larger causing the alpha particle to rebound so the atoms nucleus must be tiny but have a lot of mass.
Hello, thank you for a succint answer.

I was wondering however, were not the major deflections due to some of the nuclei having head on collisions with the gold nuclei?

So I guess we could say most of the nuclei passed straight through as the atom is composed of 99.9% empty space, and almost all of its mass is concentrated at the centre of the atom. Hence those nuclei close to the positively charged gold nuclei, will have been repelled- causing their paths to change course(or be slightly or largely deflected). Thank you for this "A curved flourescent screen was placed behind which would flash when hit by an alpha particle".

A question: Was there any need to conduct the experiment in an evacuated chamber?

Is it possible that the particles of the medium(air) would have altered the results, due to the deflection of the alpha particles caused by the air particles?
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sulexk
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It had to be performed in a vacuum because the air would absorb the alpha particles before they hit the foil or before they got to the screen!
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Oh my Ms. Coffey
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(Original post by sulexk)
Hello, thank you for a succint answer.

I was wondering however, were not the major deflections due to some of the nuclei having head on collisions with the gold nuclei?

So I guess we could say most of the nuclei passed straight through as the atom is composed of 99.9% empty space, and almost all of its mass is concentrated at the centre of the atom. Hence those nuclei close to the positively charged gold nuclei, will have been repelled- causing their paths to change course(or be slightly or largely deflected). Thank you for this "A curved flourescent screen was placed behind which would flash when hit by an alpha particle".

A question: Was there any need to conduct the experiment in an evacuated chamber?

Is it possible that the particles of the medium(air) would have altered the results, due to the deflection of the alpha particles caused by the air particles?
Yes thats the case.
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sulexk
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(Original post by Oh my Ms. Coffey)
Yes thats the case.
I have found some more information regarding what we can conclude from the alpha particle scattering experiment:

1)As most of the alpha particles passed through the gold foil, without any deflection, most of the space within the atom is empty.

2)As some of the alpha particles were deflected by large angles, they must have approached some positively charged region responsible for the deflection. This positively charged region is called the nucleus.

3) As very few alpha particles experience the deflection, it was concluded that the volume occupied by the nucleus is very small

Thank you
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anshul95
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(Original post by sulexk)
Hello, thank you for a succint answer.

I was wondering however, were not the major deflections due to some of the nuclei having head on collisions with the gold nuclei?

So I guess we could say most of the nuclei passed straight through as the atom is composed of 99.9% empty space, and almost all of its mass is concentrated at the centre of the atom. Hence those nuclei close to the positively charged gold nuclei, will have been repelled- causing their paths to change course(or be slightly or largely deflected). Thank you for this "A curved flourescent screen was placed behind which would flash when hit by an alpha particle".

A question: Was there any need to conduct the experiment in an evacuated chamber?

Is it possible that the particles of the medium(air) would have altered the results, due to the deflection of the alpha particles caused by the air particles?
as the previous poster has said you are correct. However, you must be really carefull when describing what happens when the alpha particle approaches the gold nucleus. As you correctly described, it is repelled and deflected by the nucleus. However, you also said that they have a head on collision, which would need to an insane amount of energy. I might just being picky, but if you contradict yourself when answering the questions apparently you loose loads of marks.
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ChoYunEL
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X-rays
a) describe the nature of X-rays;
(b) describe in simple terms how X-rays are produced;
(c) describe how X-rays interact with matter (limited to photoelectric
effect, Compton Effect and pair production);
(d) define intensity as the power per unit cross sectional area;
(e) select and use the equation I = I 0 e-?x to show how the intensity
I of a collimated X-ray beam varies with thickness x of medium;
(f) describe the use of X-rays in imaging internal body structures
including the use of image intensifiers and of contrast media;
(g) explain how soft tissues like the intestines can be imaged using
barium meal;
(h) describe the operation of a computerised axial tomography
(CAT) scanner;
(i) describe the advantages of a CAT scan compared with an X-ray
image.
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hannah721
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#89
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Do we need to know the energy of radioactive emissions? (page 158 of the pink ocr book)
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TobeTheHero
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#90
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(Original post by ChoYunEL)
X-rays
a) describe the nature of X-rays;

high freq, low wavelength and travel with speed of light in vacuum.

(b) describe in simple terms how X-rays are produced;

sudden deceleration of high speed electrons at the anode side

(c) describe how X-rays interact with matter (limited to photoelectric
effect, Compton Effect and pair production);


Photoelectric Effect

x-trays can interact with electrons within metal surfaces and transfer a fixed amount of energy(work function). Each ray of photon of x-ray can only interact with with one electron.

Compton Effect

An X-ray transfers some of its energy to an atomic electron the remaining energy appears as an X-ray photon, travelling in a different direction. The deflected x-ray photon has a longer wavelength as a result of travelling through a large angle( law of conservation of Momentum)

Pair Production

An X-ray strikes an atomic nucleus, producing an electron-position pair.

Energy of x-ray photon= Position + electron

this interaction can only be achieved through high voltage systems( or with high energy gamma rays).



(d) define intensity as the power per unit cross sectional area
(e) select and use the equation I = I 0 e-?x to show how the intensity
I of a collimated X-ray beam varies with thickness x of medium;

(f) describe the use of X-rays in imaging internal body structures
including the use of image intensifiers and of contrast media;


X-ray can be absorbed by bones and not absorbed by flesh, which means they can be used to examine broken bones inside the body without having to cut open the body. Contrast media used to distinguish between organs in the body, they have high atomic Z e.g. Barium meal.

(g) explain how soft tissues like the intestines can be imaged using
barium meal;

(h) describe the operation of a computerised axial tomography
(CAT) scanner;

1) Point source of x-ray
2) fan-shaped slice of x-ray produced
3) X-ray tube rotates around the position of patient
4) Fan of x-rays rotating around patient.
5) stationary ring of 720 detectors collect the beam after going through the patient.



(i) describe the advantages of a CAT scan compared with an X-ray
image.


1) they can be taken quickly, so large number of patients can be scanned per day.
2) Image produced is 3D compared X-ray(2D)
3) Low cost of running the machine.
that's what i remember
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ChoYunEL
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(Original post by TobeTheHero)
that's what i remember
5.4.2 Diagnosis methods in medicine
(a) describe the use of medical tracers like technetium-99m to
diagnose the function of organs;
(b) describe the main components of a gamma camera;
(c) describe the principles of positron emission tomography (PET);
(d) outline the principles of magnetic resonance, with reference to
precession of nuclei, Larmor frequency, resonance and relaxation
times;
(e) describe the main components of an MRI scanner;
(f) outline the use of MRI (magnetic resonance imaging) to obtain
diagnostic information about internal organs;
(g) describe the advantages and disadvantages of MRI;
(h) describe the need for non-invasive techniques in diagnosis;
(i) explain what is meant by the Doppler effect;
(j) explain qualitatively how the Doppler effect can be used to
determine the speed of blood.

How about those one?
Nicely summed up eariler - the one you didn't answer is on the formula booket then just applied. And I think you already said what (g) was in the previous statement.
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anshul95
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#92
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(Original post by TobeTheHero)
that's what i remember
only one of your advantages at the end adresses the question - low cost of running the machine and large amount of images taken per day are compared to MRI scans not compared to a normal 2D-Xray. The advantages are that an organ can be viewed from many different view points (as you can look at individual slices or rotated slices/organs) and it gives greater detail (as it you build up negative 3d images and do lots of other things that you can't really do with a convential 2d X-ray).
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Oh my Ms. Coffey
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#93
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(Original post by yokabasha)
I really need help with this, can someone explain to me how to calculate the energy released per kilogram for example 4 hydrogen atoms fusing to make a helium atom.

How would you calculate the energy released per kilogram of Hydrogen (1proton), given the total energy released in the reaction is 4.8 x 10^-12 J.

We have 1kg of hydrogen and 1 fusion reaction gives us x amount of energy per atom so we need to find the number of atoms in 1kg of hydrogen.


We can use N=(mass/atomic mass)Na to find the number of atoms in 1kg hydrogen.

I think you multiply this 4.8 x 10^-12 J by the number of atoms then divide by 4 for the number of atoms per reaction, can someone confirm this?



Per atom you have 1.2x10^-12 J as 4 atoms make the 4.8 x 10^-12 J, Multiply this by number of atoms.

I get total energy out as roughly 7.2x10^14 J
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anshul95
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#94
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(Original post by ChoYunEL)
5.4.2 Diagnosis methods in medicine
(a) describe the use of medical tracers like technetium-99m to
diagnose the function of organs;
(b) describe the main components of a gamma camera;
(c) describe the principles of positron emission tomography (PET);
(d) outline the principles of magnetic resonance, with reference to
precession of nuclei, Larmor frequency, resonance and relaxation
times;
(e) describe the main components of an MRI scanner;
(f) outline the use of MRI (magnetic resonance imaging) to obtain
diagnostic information about internal organs;
(g) describe the advantages and disadvantages of MRI;
(h) describe the need for non-invasive techniques in diagnosis;
(i) explain what is meant by the Doppler effect;
(j) explain qualitatively how the Doppler effect can be used to
determine the speed of blood.

How about those one?
Nicely summed up eariler - the one you didn't answer is on the formula booket then just applied. And I think you already said what (g) was in the previous statement.
a)- medical tracers are used to give metabolic information about the body. They can show whether an organ is absorbing a substance properly/abnormal metabolic rate. Medical tracers are either attatched to a molecule such as glucose where they are absorbed by organs or a radioactive element of a molecule which is normally absorbed by the organ is used (e.g. iodine to look at thyroid function). Technetium is suitable as it emits gamma rays (and therefore doesn't ionise much surrounding cells in the body). It has a faiirly short half life so that it doesn't expose the patient to ionising radiation for too long but long enough to allow for time taken to diagnose and form transportation of tracer.

b)gamma camera has a collimator which only lets vertical gamma photons through. A scintillator to transfer gamma ray photons to visible light photons. A photomultiplier tube connected to a computer which converts the light into an electrical signal to be interpreted by a computer.

c)a positron emitting substance is attatched to a molecule such as glucose. When absorbed by the body, the positron emitting substance emits a positron which annihilates with a surrounding electron producing two gamma rays in opposite directions. A ring of detectors surrounds (after some time is allowed for the substance to be absorbed properly) the body which, if it detects two simulataneous photons, inteprets this as the annhilation occuring along the line between the detectors. The information is then sent to a computer for display.

d) the hydrogen nuclei have spin which means they have magnetic moment (as they have unequal amount of protons and neutrons). The hydrogen nuclei normally spin on their own axis. When exposed to a strong external magnetic field, these nuclei precess (they rotate in a circle in the horizontal plane). When RF radiation of a particular frequency is sent through, resonance occurs (lamour freq) and the nuclei precesses against the direction of the external magnetic field (high energy state). Once RF transmission is stopped, the nuclei flip back to the low energy state, releasing the difference in energy as another RF wave of the same frequency. The time taken for nuclei to travel from high energy to low energy state is the relaxation time and is affected by the hydrogen environments i.e. surrounding tissue types.

e)The main magnets produce the strong external magnetic field required by using superconduction. The additional magnets vary the lamour freq by providing a gradient field in 3D space so that the precise location of relaxations can be calculated. The RF coils produces short pulses of RF. The waves are pulsed so that the start of each relaxation is known and short to make sure the emitted and received RF radiation to not overlap and interferre.

f)MRI can be used to be distinguish between many different tissue types such as fats, cancerous tissues etc.

g) adv- high image quality. Can distinguish between many different tissue types. No ionising radiation is used.

disadv- expensive due to supercooled magnets. Can interferre with electronic pace makers/any metallic object. Takes a long time to scan one patient so less scans can be taken each day.

h) With the exception of MRI they are often quicker. Can be used especially with pregnant women/patients who at risk from other treatments. Probably not a good answer for this.

i) Doppler effect is the shifting in wavelength when there is relative velocity between the wave source and the observer.

j) Well the change in frequency of the reflected ultrasound can be used to work out the velocity of blood moving through arteries and veins which can be used to diagnose arterial diseases etc.
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anshul95
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#95
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(Original post by Oh my Ms. Coffey)
We have 1kg of hydrogen and 1 fusion reaction gives us x amount of energy per atom so we need to find the number of atoms in 1kg of hydrogen.


We can use N=(mass/atomic mass)Na to find the number of atoms in 1kg hydrogen.

I think you multiply this 4.8 x 10^-12 J by the number of atoms then divide by 4 for the number of atoms per reaction, can someone confirm this?



Per atom you have 1.2x10^-12 J as 4 atoms make the 4.8 x 10^-12 J, Multiply this by number of atoms.

I get total energy out as roughly 7.2x10^14 J
I am not sure I am following your reasoning. The correct answer is
(4.8*10^-12*1000*6.02*10^23)/4

Edit: I have just seen your end answer yes you are correct as I got the same answer so we must have used the same method.
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Pheylan
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#96
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6.69*10^(-27) kg = 4.8*10^(-12) J

so 1 kg = 7.14*10^(14) J
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Oh my Ms. Coffey
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(Original post by anshul95)
I am not sure I am following your reasoning. The correct answer is
(4.8*10^-12*1000*6.02*10^23)/4

Edit: I have just seen your end answer yes you are correct as I got the same answer so we must have used the same method.
Yeah sorry it was a bit long winded, I prefer writing it up on paper.

Number of hydrogen atoms in 1k, N=(1/0.001)6.02x10^23 = 6.02x10^26.

Per reaction 4.8x10^-12 J released when 4H converts to 1He so the energy when 1 hydrogen atom fuses is 1.2x10^-12 J.

Energy in the fusion of 1KG Hydrogen = Energy per hydrogen atom x Number of atoms.

1.2x10^-12 x 6.02x10^-26 = 7.224x10^14 J.
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Ralphus J
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#98
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I really need help on the understanding of mass spectrometry

Thanks
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PhilliChilli
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#99
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Can anyone summarise the formation of a star and how neutron stars are made?
I keep getting this wrong and apparently missing out a lot of keywords :/
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ChoYunEL
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(Original post by Ralphus J)
I really need help on the understanding of mass spectrometry

Thanks
A Mass Spectrometry uses the principle of f = (m*v^2)/r
The Spectrometer will project/accelerate a substance of unknown material/substances through a magnetic and/or Electric field.
Force/Velocity^squared = Mass/Radius

Since you cannot change the mass of an material, a detector/collector picks up where the material is using the radius it has traveled through the magnetic field.

There are two types of mass spectrometer as far as I understand..
One which only allow one particular substance through a gap (using electric fields to stop additional ions)
And ones which has several detectors, without an electric field, which collects certain irons.

Please correct me if I am wrong.
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