Presentation on theme: "X-RAYS In 1895 at the University of Wurzburg, Wilhelm Roentgen (1845–1923) was studying electrical discharges in low-pressure gases when he noticed that."— Presentation transcript:
X-RAYS In 1895 at the University of Wurzburg, Wilhelm Roentgen (1845–1923) was studying electrical discharges in low-pressure gases when he noticed that a fluorescent screen glowed even when placed several meters from the gas discharge tube and even when black cardboard was placed between the tube and the screen He concluded that the effect was caused by a mysterious type of radiation, which he called x-rays because of their unknown nature. Subsequent study showed that these rays travelled at or near the speed of light and that they couldn’t be deflected by either electric or magnetic fields.
In 1912 Max von Laue demonstrated that such a diffraction pattern could be observed, similar to that shown in Figure 27.7 for NaCl.X-rays are produced when high-speed electrons are suddenly slowed down— for example, when a metal target is struck by electrons that have been accelerated through a potential difference of several thousand volts. A current in the filament causes electrons to be emitted, and these freed electrons are accelerated toward a dense metal target, such as tungsten, which is held at a higher potential than the filament.
The first step in the production of characteristic x-rays occurs when a bombarding electron collides with an electron in an inner shell of a target atom with sufficient energy to remove the electron from the atom. The vacancy created in the shell is filled when an electron in a higher level drops down into the lower energy level containing the vacancy. The time it takes for this to happen is very short, less than s. The transition is accompanied by the emission of a photon with energy equalling the difference in energy between the two levels. Typically, the energy of such transitions is greater than eV, and the emitted x-ray photons have wavelengths in the range of 0.01 nm to 1 nm.
Note that the spectrum has are two distinct components. One component is a continuous broad spectrum that depends on the voltage applied to the tube. Superimposed on this component is a series of sharp, intense lines that depend on the nature of the target material. The continuous radiation is sometimes called bremsstrahlung Because the radiated photon shown in Figure carries energy, the electron must lose kinetic energy because of its encounter with the target.
An extreme example would consist of the electron losing all of its energy in a single collision. In this case, the initial energy of the electron (eΔV ) is transformed completely into the energy of the photon (hf max ). In equation form where eΔV is the energy of the electron after it has been accelerated through a potential difference of ΔV volts and e is the charge on the electron. This says that the shortest wavelength radiation that can be produced is Interesting insights into the process of painting and revising a masterpiece are being revealed by x-rays.
EXAMPLE: Medical x-ray machines typically operate at a potential difference of 1.00 x10 5 V. Calculate the minimum wavelength their x-ray tubes produce when electrons are accelerated through this potential difference. Solution Substitute into Equation 27.9:
27.5 THE COMPTON EFFECT Arthur H. Compton in 1923 directed an x-ray beam of wavelength λ 0 toward a block of graphite. He found that the scattered x-rays had a slightly longer wavelength λ than the incident x-rays, and hence the energies of the scattered rays were lower. The amount of energy reduction depended on the angle at which the x-rays were scattered. The change in wavelength Δλ between a scattered x-ray and an incident x-ray is called the Compton shift.
In order to explain this effect, Compton assumed that if a photon behaves like a particle, its collision with other particles is similar to a collision between two billiard balls. Hence, the x-ray photon carries both measurable energy and momentum, and these two quantities must be conserved in a collision. If the incident photon collides with an electron initially at rest, as in Figure 27.16, the photon transfers some of its energy and momentum to the electron. As a consequence, the energy and frequency of the scattered photon are lowered and its wavelength increases.
Applying relativistic energy and momentum conservation to the collision described in Figure 27.16, the shift in wavelength of the scattered photon is given by where m e is the mass of the electron and θ is the angle between the directions of the scattered and incident photons. The quantity h/m e c is called the Compton wavelength and has a value of nm. The Compton wavelength is very small relative to the wavelengths of visible light, so the shift in wavelength would be difficult to detect if visible light were used
EXAMPLE: X-rays of wavelength λ 0 = nm are scattered from a block of material. The scattered x-rays are observed at an angle of 45.0° to the incident beam. (a) Calculate the wavelength of the x-rays scattered at this angle. (b) Compute the fractional change in the energy of a photon in the collision. Solution Calculate the wavelength of the x-rays. Substitute into Equation to obtain the wavelength shift:
Add this shift to the original wavelength to obtain the wavelength of the scattered photon: Find the fraction of energy lost by the photon in the collision. Rewrite the energy E in terms of wavelength, using c = fλ: Compute ΔE/E using this expression: Cancel hc and rearrange terms: Substitute values from part (a):
THE DUAL NATURE OF LIGHT AND MATTER a)Light and Electromagnetic Radiation Phenomena such as the photoelectric effect and the Compton Effect offer evidence that when light and matter interact, the light behaves as if it were composed of particles having energy hf and momentum h/λ. In other contexts, however, light acts like a wave, exhibiting interference and diffraction effects. Is light a wave or a particle? The answer depends on the phenomenon being observed Light has a dual nature, exhibiting both wave and particle characteristics.
In low frequencies the energy of a photon having this frequency is only about eV, too small to allow the photon to be detected we use antenna to be able to detect the individual photons striking the antenna. As we go to higher frequencies in the visible region, it’s possible to observe both the particle characteristics and the wave characteristics of light. a light beam shows interference phenomena (thus, it is a wave) and at the same time can produce photoelectrons (thus, it is a particle). At even higher frequencies, the momentum and energy of the photons increase. Consequently, the particle nature of light becomes more evident than its wave nature.
The Wave Properties of Particles In his doctoral dissertation in 1924, Louis de Broglie postulated that, because photons have wave and particle characteristics, perhaps all forms of matter have both properties. According to de Broglie, electrons, just like light, have a dual particle–wave nature. The relationship between energy and momentum for a photon, which has a rest energy of zero, is p = E/c. We also know from Equation 27.5 that the energy of a photon is Consequently, the momentum of a photon can be expressed as
From this equation, we see that the photon wavelength can be specified by its momentum, or λ= h/p. De Broglie suggested that all material particles with momentum p should have a characteristic wavelength λ = h/p. Because the momentum of a particle of mass m and speed v is mv = p, the de Broglie wavelength of a particle is de Broglie postulated that the frequencies of matter waves (waves associated with particles having nonzero rest energy) obey the Einstein relationship for photons, E = hf, so that The fact that these relationships had been established experimentally for photons made the de Broglie hypothesis that much easier to accept.
EXAMPLE: Compare the de Broglie wavelength for an electron (m e = 9.11 x kg) moving at a speed of 1.00 x 10 7 m/s with that of a baseball of mass kg pitched at 45.0 m/s. (b) Compare these wavelengths with that of an electron travelling at 0.999c. Solution Compare the de Broglie wavelengths of the electron and the baseball. Substitute data for the electron into Equation 27.14:
Repeat the calculation with the baseball data: Find the wavelength for an electron traveling at 0.999c.Replace the momentum in Equation with the relativistic momentum: Substitute:
The electron wavelength corresponds to that of x-rays in the electromagnetic spectrum. The baseball, by contrast, has a wavelength much smaller than any aperture through which the baseball could possibly pass, so we couldn’t observe any of its diffraction effects. It is generally true that the wave properties of large-scale objects can’t be observed. Notice that even at extreme relativistic speeds, the electron wavelength is still far larger than the baseball’s.
27.7 THE UNCERTAINTY PRINCIPLE According to classical mechanics, no fundamental barrier to an ultimate refinement of the apparatus or experimental procedures exists. In other words, it’s possible, in principle, to make such measurements with arbitrarily small uncertainty. Quantum theory predicts, however, that such a barrier does exist. 1927, Werner Heisenberg (1901–1976) introduced this notion, which is now known as the uncertainty principle:
In other words, it is physically impossible to measure simultaneously the exact position and exact linear momentum of a particle. If Δx is very small, then Δp x is large, and vice versa. Thus, in the process of locating the electron very accurately (that is, by making Δx very small), the light that enables you to succeed in your measurement changes the electron’s momentum to some undeterminable extent (making Δp x very large). The incoming photon has momentum h/λ. As a result of the collision, the photon transfers part or all of its momentum along the x-axis to the electron.
Therefore, the uncertainty in the electron’s momentum after the collision is as great as the momentum of the incoming photon:Δp x = h/λ. Further, because the photon also has wave properties, we expect to be able to determine the electron’s position to within one wavelength of the light being used to view it, so Δx = λ. Multiplying these two uncertainties gives The value h represents the minimum in the product of the uncertainties. Because the uncertainty can always be greater than this minimum, we have
Apart from the numerical factor 1/4π introduced by Heisenberg’s more precise analysis, this inequality agrees with Equation Another form of the uncertainty relationship sets a limit on the accuracy with which the energy E of a system can be measured in a finite time interval Δt : It can be inferred from this relationship that the energy of a particle cannot be measured with complete precision in a very short interval of time. Thus, when an electron is viewed as a particle, the uncertainty principle tells us that (a) its position and velocity cannot both be known precisely at the same time and (b) its energy can be uncertain for a period given by Δt = h/(4πΔE ).
EXAMPLE: The speed of an electron is measured to be 5.00 x 10 3 m/s to an accuracy of %. Find the minimum uncertainty in determining the position of this electron. Solution Calculate the momentum of the electron: The uncertainty in p is % of this value: Now calculate the uncertainty in position using this value of Δp x and Equation 27.17:
EXAMPLE: As we’ll see in the next chapter, electrons in atoms can be found in certain high states of energy called excited states for short periods of time. If the average time that an electron exists in one of these states is 1.00x s, what is the minimum uncertainty in energy of the excited state? Solution Use Equation to obtain the minimum uncertainty in the energy: This is again an imprecise calculation, giving only a lower bound on the uncertainty.