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# Physics Calculations: Radiography And Tomography (Math Problem Sample)

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Physics calculations

source..Content:

Radiography and Tomography

By [Name of Student]

Course Name

[Name of Affiliate Institution]

Professor’s Name

Date

1. What means energy will one get when the X-ray machine is set to 200kV? Explain your answer in no less than 400 words and attach a figure(s) where applicable. [15]

The mean energy of the X-ray machine for 200kV, when the tube current is 400mA is 80kW of energy produced or accelerated. The spectrum for the 200 kV beam is seen as shown in the figure below consists of a broad Bremsstrahlung spectrum with superimposed K-characteristic lines but this time low energy X-rays is filtered. Therefore, the mean X-ray beam is raised through filtration. The intensity of energy at 200 kV is sufficient to generate K- characteristics Radiation in the atoms of tungsten.

The diagram shows that the mean energy of X-ray is located where the intensity of the 200kV is discontinued. The amount of voltage at that point is the measure of the intensity of the X-ray protons radiation.

The photons generated by the number of X-rays at the anode end generally depends on the amount of energy which is the product of current and Voltage. Therefore, the impact of mA the X-ray energy spread is raising the energies and at the same time, the spread or spectrum remain intact. Note that it is the product of tube in the recorded radiography, the mean energy of X-ray is calculated by current (mA) and time of exposure (s). The amount of energy, Emean, generated from 200kV:

Emean = kVmean x mAmean x exposure time.

The expression of the average energy of x-ray depends on the measured time of exposure which can be in second thus energy in joules. For this radiography, approximately amount of energy generated is about 250 kJ in general. This amount of energy is obtained from the product of the three functions; Voltage, current and Exposure time. The amount for the model is generally at mean energy. However, there is still allowance for minimum and maximum energy that this radiography X-ray machine can produce.

The amount of X-ray energy generated at the anode largely depends on the amount of the voltage applied which has the relationship of about square i.e. the energy production is directly proportional to the square of the voltage . Therefore, the X-ray production energy at 200kV is substantially higher. In the figure below, it shows the relationship of the X-ray energy produced in terms of protons emission where at about 200kV, the amount of photons emitted is about 20 times the Voltage amount. Thus, the relationship of X-ray energy spread majorly relies on the amount of kV which determines the X-ray protons in the spectrum of the beam.

2. Assuming that the setup has a geometric unsharpness of 4, a focal spot of 1.1 and object-detector distance of 150cm. What is the magnification of the setup? [5]

Geometric unsharpness or the size of the penumbra (U) is obtained using the following equation:

U=f×ab

Where: f = X-ray generator focal-spot size. a = distance from the x-ray source to the front surface of material/object b = distance from the material/object to the detector

Using the known parameters; U= 4, f= 1.1 and b=150cm. Hence;

a=Ubf=4×1501.1=545.45 cm

magnification=ab=545.45150=3.64

3. You have many camera options in your laboratory. A client requires to measure a sphere with diameter 40cm (horizontal length and vertical length) and wants to measure an entire object to an OR of 15 microns/pixel. What CCD chip number of pixels will you require? [3]

1cm= 10000 micron; 40 cm will need; 40×10000=400,000 microns

And 1 pixel = 15 microns hence;

40000015=2666.67 CCD chip of pixels

4. If you have a camera with which you want to image with the resolution 2 micrometers. What should be the distance between the lens and the object you are imaging if the lens radius 1.5 meters, wavelength 7 nanometers? [6]

Hint. Use Rayleigh criterion.

θ=1.22ƛD=xd

Where θ is aperture angle, ƛ is wave length, D is dthe iameter of lthe ens, x is resolution distance and d distance between the lens and object

Rewriting the formula,

x=1.22ƛdD=2×10-3=1.227×10-9×d3

d=2×10-3×31.22×7×10-9=7.026 ×105m

5. Derive the equation for Spectral Signal-Noise Ratio (SSRN)? [20]

In the derivation of SSNR equation, the case of the non-zero mean is considered where statistically, the noise covariant is given by;

∈m∈n=σr,n2δm,n…………………………………….1

Where δm,n is delta function of Kronecker

σr,n2=σd2+hn∆τ……………………………………2

σd2 is the detector noise variance and hτ is the short noise variance that is linked to pattern intensity interference and is in the detected photoelectrons unit.

Typical approximation which is valid;

σr,n2≈σd2+h=σd2+td20∞Svdv…………………3

Taking an assumption that in each independent sample of noise, the real part variance of the DFT of the noise is calculated by;

σr,p2=1Nn=-N2N2∈mcos-i2πpnN2

=1Nn=-N2N2σr,n2cos22πpnN………………..4

By using equations 1 and 2 to substitute in the equation 4, which yields the simplified equation below;

σr,p2=σd22+td40∞Svdv=1+δp,0+δp,-N2………5

The SSNR is therefore defined as the ratio of spectral signal Gp to the deviation standard of noise σr,p2 which is:

SSNR=Gp σr,p2………………………………….6

In detector noise limit, when there is the dominance of noise that is;

σd2≫td20∞Svdv……………………..7

Then the SSNR can be estimated as:

SSNR=tdq=-∞∞Sp-qN∆v+S-p-qN∆v22Nσd∆τ1+δp,0+δp,-N2………..8

SSNR∝tdN∆τ=Ntd∆τ……………………….9

When short noise is limited;

td20∞Svdv≫σd2……………………..10

Then the SSNR can be estimated as:

SSNR=tdq=-∞∞Sp-qN∆v+S-p-qN∆v22N∆τ0∞Svdv1+δp,0+δp,-N2………..11

SSNR∝tdN∆τ=Ntd∆τ……………………….12

Hence the derived equation of SSNR is shown by equation 9 and 12.

6. Discuss in detail neutron radiography using polarised neutrons. This should include the neutron properties and the detection system. [15]

The process of producing or generating neutron image which is film recorded is called neutron radiography. The method is with the highest resolution which is competitive with digital setups. Through the Zeeman interaction, the neutrons polarization particles are spin ½ which interacts with the field of magnetic induction. In polarization, the procession technique used in induction field is:

dMdt=γnM&tim...

By [Name of Student]

Course Name

[Name of Affiliate Institution]

Professor’s Name

Date

1. What means energy will one get when the X-ray machine is set to 200kV? Explain your answer in no less than 400 words and attach a figure(s) where applicable. [15]

The mean energy of the X-ray machine for 200kV, when the tube current is 400mA is 80kW of energy produced or accelerated. The spectrum for the 200 kV beam is seen as shown in the figure below consists of a broad Bremsstrahlung spectrum with superimposed K-characteristic lines but this time low energy X-rays is filtered. Therefore, the mean X-ray beam is raised through filtration. The intensity of energy at 200 kV is sufficient to generate K- characteristics Radiation in the atoms of tungsten.

The diagram shows that the mean energy of X-ray is located where the intensity of the 200kV is discontinued. The amount of voltage at that point is the measure of the intensity of the X-ray protons radiation.

The photons generated by the number of X-rays at the anode end generally depends on the amount of energy which is the product of current and Voltage. Therefore, the impact of mA the X-ray energy spread is raising the energies and at the same time, the spread or spectrum remain intact. Note that it is the product of tube in the recorded radiography, the mean energy of X-ray is calculated by current (mA) and time of exposure (s). The amount of energy, Emean, generated from 200kV:

Emean = kVmean x mAmean x exposure time.

The expression of the average energy of x-ray depends on the measured time of exposure which can be in second thus energy in joules. For this radiography, approximately amount of energy generated is about 250 kJ in general. This amount of energy is obtained from the product of the three functions; Voltage, current and Exposure time. The amount for the model is generally at mean energy. However, there is still allowance for minimum and maximum energy that this radiography X-ray machine can produce.

The amount of X-ray energy generated at the anode largely depends on the amount of the voltage applied which has the relationship of about square i.e. the energy production is directly proportional to the square of the voltage . Therefore, the X-ray production energy at 200kV is substantially higher. In the figure below, it shows the relationship of the X-ray energy produced in terms of protons emission where at about 200kV, the amount of photons emitted is about 20 times the Voltage amount. Thus, the relationship of X-ray energy spread majorly relies on the amount of kV which determines the X-ray protons in the spectrum of the beam.

2. Assuming that the setup has a geometric unsharpness of 4, a focal spot of 1.1 and object-detector distance of 150cm. What is the magnification of the setup? [5]

Geometric unsharpness or the size of the penumbra (U) is obtained using the following equation:

U=f×ab

Where: f = X-ray generator focal-spot size. a = distance from the x-ray source to the front surface of material/object b = distance from the material/object to the detector

Using the known parameters; U= 4, f= 1.1 and b=150cm. Hence;

a=Ubf=4×1501.1=545.45 cm

magnification=ab=545.45150=3.64

3. You have many camera options in your laboratory. A client requires to measure a sphere with diameter 40cm (horizontal length and vertical length) and wants to measure an entire object to an OR of 15 microns/pixel. What CCD chip number of pixels will you require? [3]

1cm= 10000 micron; 40 cm will need; 40×10000=400,000 microns

And 1 pixel = 15 microns hence;

40000015=2666.67 CCD chip of pixels

4. If you have a camera with which you want to image with the resolution 2 micrometers. What should be the distance between the lens and the object you are imaging if the lens radius 1.5 meters, wavelength 7 nanometers? [6]

Hint. Use Rayleigh criterion.

θ=1.22ƛD=xd

Where θ is aperture angle, ƛ is wave length, D is dthe iameter of lthe ens, x is resolution distance and d distance between the lens and object

Rewriting the formula,

x=1.22ƛdD=2×10-3=1.227×10-9×d3

d=2×10-3×31.22×7×10-9=7.026 ×105m

5. Derive the equation for Spectral Signal-Noise Ratio (SSRN)? [20]

In the derivation of SSNR equation, the case of the non-zero mean is considered where statistically, the noise covariant is given by;

∈m∈n=σr,n2δm,n…………………………………….1

Where δm,n is delta function of Kronecker

σr,n2=σd2+hn∆τ……………………………………2

σd2 is the detector noise variance and hτ is the short noise variance that is linked to pattern intensity interference and is in the detected photoelectrons unit.

Typical approximation which is valid;

σr,n2≈σd2+h=σd2+td20∞Svdv…………………3

Taking an assumption that in each independent sample of noise, the real part variance of the DFT of the noise is calculated by;

σr,p2=1Nn=-N2N2∈mcos-i2πpnN2

=1Nn=-N2N2σr,n2cos22πpnN………………..4

By using equations 1 and 2 to substitute in the equation 4, which yields the simplified equation below;

σr,p2=σd22+td40∞Svdv=1+δp,0+δp,-N2………5

The SSNR is therefore defined as the ratio of spectral signal Gp to the deviation standard of noise σr,p2 which is:

SSNR=Gp σr,p2………………………………….6

In detector noise limit, when there is the dominance of noise that is;

σd2≫td20∞Svdv……………………..7

Then the SSNR can be estimated as:

SSNR=tdq=-∞∞Sp-qN∆v+S-p-qN∆v22Nσd∆τ1+δp,0+δp,-N2………..8

SSNR∝tdN∆τ=Ntd∆τ……………………….9

When short noise is limited;

td20∞Svdv≫σd2……………………..10

Then the SSNR can be estimated as:

SSNR=tdq=-∞∞Sp-qN∆v+S-p-qN∆v22N∆τ0∞Svdv1+δp,0+δp,-N2………..11

SSNR∝tdN∆τ=Ntd∆τ……………………….12

Hence the derived equation of SSNR is shown by equation 9 and 12.

6. Discuss in detail neutron radiography using polarised neutrons. This should include the neutron properties and the detection system. [15]

The process of producing or generating neutron image which is film recorded is called neutron radiography. The method is with the highest resolution which is competitive with digital setups. Through the Zeeman interaction, the neutrons polarization particles are spin ½ which interacts with the field of magnetic induction. In polarization, the procession technique used in induction field is:

dMdt=γnM&tim...

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