大学物理实验报告英文版X光

1、experiment report of x-ray principle:when we speak of attenuation of x-rays, we mean the decrease in intensity that occurs when the radiation passes through matter. this attenuation is caused mainly by the two effects: scattering and absorption.although absorption and attenuation are different physi

2、cal phenomena, the transilluminated object is often referred to -inaccurately- as an absorber; this should more properly be termed an attenuator. however, this description will follow the traditional usage in some places and refer to absorbers instead of attenuators.the scattering of x-ray quanta at

3、 the atoms of the attenuator material causes a part of the radiation to change direction. this reduces the intensity in original direction. this scattering can be either elastic or entail an energy loss or shift in wavelength, i.e. inelastic scattering.when x-ray hits a fluorescence material, it exc

4、ites the material, making fluorescence lights. this is the principle of a fluorescence screen. when x-ray is used in imaging, the transmitted light will carry interior information of the object. a more sophisticated x-ray detector is the so-called geiger-muller counter, an instrument for detecting t

5、he presence of and measuring ionizing radiation such as the x-rays. it converts the intensity of the x-ray into counting rates. we will use such a device in our lab.attenuation of x-rays:when passing through a material, x-ray can be attenuated by e&m interactions. for a slab of thickness x, the

6、transmission (defined as the ratio of the transmitted radiation to the incoming), . µ is the attenuation coefficient, with a dimension of 1/distance. µis a character of the material, and it varies, for example, as a function of atomic number. we will study this dependence in this lab.bragg

7、 diffractionlike normal lights, when x-ray transmit through material with regular optical pattern (e.g. lattice), diffraction will happen if the wavelength of the x-ray is close to the lattice space. such diffraction on the crystal is the so-called bragg diffraction. if the lattice spacing is d, and

8、 the x-ray and the crystal surface forms an angle q, the angle where maximum diffraction happens will satisfy where n is an integer and l is the wavelength. lab equipment: the equipment is an x-ray lab system made by leybold inc. a schematic is shown below.the x-ray is generated by an electron beam

9、with controllable energy (via the potential) and current. the x-ray is going into the detection chamber to the right. there is a removable aperture which focuses the x-ray, a rotatable sample holder, and a rotatable g-m counter. at the right end of the wall, there is a fluorescence screen for imagin

10、e. operation details of the device will be given by the manual. basically, you need to set the high voltage (u) which determine the energy of the x-ray, the current (i), and the angle of the sample holder (target) or the detector. a knob can be used to make the adjustment on selected parameters. “co

11、upled” movement means one moves the target and the detector together, the former by an angle a, and the latter by an angle 2a. make sure the lead glass window is closed before you turn on the high voltage.attenuator (target)left: aluminum attenuator mounted on a curved plate with thickness of 0，0.5,

12、1.0,1.5,2.0,2.5, and 3mm. one can select different thickness by selecting angle. right: attenuation of different materials, all with a thickness of 0.5 mm, including polystyrene (average z=6), aluminum (13), iron (26), copper (29), zirconium(40) and silver(47).2.install the aperture. install the zr

13、foil onto the aperture. this is to filter out the k line. squeeze the nacl crystal onto the sample holder.in this part of the procedure we need to make sure that the x ray beam, the crystal surface and the detector is aligned. use the following alignment procedure by the bragg diffraction, the proce

14、ss is omitted.data:the first peak appears when its 7.5 degree.1. analysis:then, m. that is, the wavelength of x ray is m.2.measure the x ray attenuation to different thickness:install the zr foil onto the aperture. set hv=21 kv, i=0.05 ma, db=0, dt=100 s.hit target key. change the angle sequentially

15、 to 0, 10, 20, 30, 40, 50, 60 degree (the thickness of the attenuation increase by 0.5mm each step).for each of this setting, hit “scan” to take the data, and hit “replay” to get the average rate.attenuation of different thicknessthickness(mm)0.511.522.53data(1/s)21.49.64.42.81.71fit the rate as a f

16、unction of thickness (with background subtraction) to get the attenuation coefficient.fit the data in origin with exponential fit (mode = expdec1), through equation. we form the graph of 1/t and x.so, we get thathence,so, the attenuation coefficient is .measure the x ray attenuation to materials wit

17、h different atomic mass:a) remove the zr filter. insert the curved attenuator holder b (different target) into the annulus slot on the plastic mount. set hv = 30 kv, i=1 ma, db=0, dt=100 s. hit target key. change the angle sequentially to 0, 10, 20, 30, 40, 50, 60 degree, each corresponding to a giv

18、en material.b) for each of this setting, hit “scan” to take the data, and hit “replay” to get the average rate.c) for the scan with zr filter.attenuation of different z, with zrz61326294047data(1/s)9117.97627.472.97.7108.914.4no exact rule can be found.install the zr filter. repeat of z number.the graph turns out:attenuation of different z, without zrz61326294047data(1/s)9025.49027.88308.475.25.274.8plot the rate as a function of z number.also,no exact rule can be found.conclusion: 1.