# MSE 104 Problem Set 1

February 5, 2018 | Author: Sukhpreet Kaur | Category: Electron, Vacuum Tube, Radiation, X Ray, Atmosphere Of Earth

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Sukhpreet Kaur 23026113 MSE 104 Problem Set 1

1.1 What is the frequency (per second) and energy per quantum(in joules) of x-ray beams of wavelength .71 ΗΊ (Mo KΞ±) and 1.5 ΗΊ (Cu KΞ±). π

The frequency of the wave is given by π = π, where f= frequency, c = velocity of light = 3.00 Γ 108 π/π ππ, and π = π€ππ£ππππππ‘β. So,

π

3.00Γ108

π

π

3.00Γ108

π

π = π = .71 Γ10β10 sπ = π. ππ Γ ππππ πβπ

And π = π = 1.5 Γ10β10sπ = π. ππ Γ ππππ πβπ Energy = E = hf where h= 6.63 Γ 10β34 πππ’πππ  β π  So, Energy of Mo= πΈ = ( 6.63 Γ 10β34 πππ’πππ  β π  ) Γ 4.23 Γ 1018 π  β1 = π. π Γ ππβππ ππππππ Energy of Cu= πΈ = ( 6.63 Γ 10β34 πππ’πππ  β π  ) Γ 1.98 Γ 1018 π  β1 = π. ππ Γ ππβππ ππππππ

1.3 Show that the velocity with which electrons strike the target of an x-ray tube depends only on the voltage between anode (target) and cathode and not on the distance between them. [The force on a charge e (coulombs) by a field E (volts/m) in eE newtons. ] The kinetic energy of the electrons on impact with the target is, πΎπΈ = ππ =

1 ππ£ 2 2

where e= charge on electron, V is the voltage, m is the mass of the electro and v is the velocity. In x-ray tube, the high voltage draws the electrons to the target/anode, which they strike with very high velocity. The tube is a vacuum tube, thus the electrons do not feel inference of any other force accept that from the voltage. Since this is true, electron will stay in constant motion and will not accelerate over distance, so variance in distance doesnβt matter. When the electrons strike the anode, x rays are produced at the point of impact.

1.6 Graphically verify Eq. (1-13) for a lead absorber and Mo KΞ±, Rh KΞ± and Ag KΞ± radiation. ( The mass absorption coefficients of lead for these radiations are 122.8, 84.13 and 66.14 cm^2/g, respectively). From the curve, determine the mass absorption coefficient of lead for the shortest wavelength radiation from a tube operated at 30,000 volts.

Equation 1-13 is: π π

= π Ξ»3 π 3

π

Where π is the mass absorption coefficient, π is the density and Z is the atomic number of the absorber. The wavelengths, along with the absorption edges ( .14 and .98) of Pb, were found in from Appendix 7. By using the absorption edge and the data given, the following graph was produced:

Mass Absorption Coefficient and Wavelength plot Mass absorption coefficient

140 120 100 80

60 40 20 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Wavelength

The curve produced is roughly similar to how the graph of the equation y^3 would look like. Thus it could be seen that the Eq 1-13 is right. The shortest wavelength limit can be found by using the following equation: ππππΏ =

1.24 Γ 103 , π€βπππ π = π£πππ‘πππ π

So, for 30,000 volts, ππππΏ =

1.24 Γ103 30000

= .4133 ΗΊ

To get the mass absorption coefficient for this wavelength, interpolate from the above mass.

π π

~ 45 cm^2 / g

1.8 (a) Calculate the mass and linear absorption coefficients of air for Cr KΞ± radiation. Assume that air contains 80% nitrogen and 20 % oxygen by weight and has a density of 1.29 Γ π 10β3 ππ^3 . (b) Plot the transmission factor of air for Cr KΞ± radiation and a path length of 0 to 20 cm. (a) Air is composed of more than one element. In this case, the mass absorption coefficient is the weighted average of the mass absorption coefficients of the two elements itβs composed of. According to 1-12. π π π = π€1 ( ) + π€2 ( ) + β― π π 1 π 2 Since air contains 80% nitrogen and 20 % oxygen, π π πππ

π

π

= (. 8) (π) + (. 2) (π) = (. 8)(24.42) + (. 2)(37.19) = 26.97 π

π

ππ2 ππ

The mass absorption values of N and O for Cr Ka radiation were found from the Appendix 8 of the textbook. The linear absorption coefficient is π = (26.97

ππ2 π

) ( 1.29 Γ

10β3 π ππ3

) = 3.48 Γ 10β2 ππβ1

(b) When x-rays come in contact with matter, they are transmitted and partly absorbed. Intensity of the x-ray beam as it passes through an object tends to decrease proportionally to the distance traveled x. The transmission factor is the ratio of final intensity , πΌπ₯ , after penetrating the length x over the initial, incident intensity, πΌ0 . Itβs given by the following equation: πΌπ₯ = π βππ₯ πΌ0 The linear absorption coefficient was calculated earlier, and by using various path lengths between 0 to 20 cm, the following graph was generated. The following graph plots the transmission factor of air in relation to the path lengths. As can be seen from the graph below, the transmission factor decreased as the path length increased.

Transmission factor of air for Cr KΞ± radiation and vs Path length Plot. 1.2

Transmission Factor

1 0.8 0.6

0.4 0.2 0 0

5

10 Path Length 15

20

25