1. Explain why H and He cannot be detected by Auger electron spectroscopy. (10 points) 2. Why are MgK? and AlK? used as X-ray sources for XPS? Do they excite 1s photoelectron emission of metals such as Ti and Fe? If not, how do we detect such elements? (10 points) 3. In XPS, we need to detect energy shifts of 0.1 eV at energy level up to 1500 eV. For an energy analyzer with the energy resolution limit, ?E/E = 5.3 x 10?³, does a pass energy of 10 eV satisfy the requirement? (10 points) 4. We may estimate the lifetime of a surface monolayer by assuming that each primary ion will knock out one atom in the monolayer. Estimate the lifetime of surface monolayer bombarded by primary ion beam densities of 10 nA cm?² and 0.1 nA cm?², respectively. Why should the primary ion dose be even less than that based on the lifetime estimation? (10 points)
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Hydrogen (H) and Helium (He) cannot be detected by Auger electron spectroscopy because they only have one electron shell. Auger electron spectroscopy works by detecting the energy of electrons emitted from an atom's inner shells when they are excited by a Show more…
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Energy dispersive spectroscopy (EDS) is a technique for identifying and quantifying the elemental composition of a sample. Volumes as small as a few cubic micrometers can be probed using EDS. The characteristic X-rays of the elements in a sample are produced when the sample is bombarded with electrons in an electron beam instrument such as a scanning electron microscope. For elements heavier than hydrogen, we can approximate the energy of an electron in the $n=1$ state (called a $K$ -shell electron) using $$E_{K}=-\frac{(13.6 \mathrm{eV})(Z-1)^{2}}{n^{2}}$$ with $n=1 .$ Here $Z$ is the atomic number of the element, and the $(Z-1)$ term corrects for the partial cancellation of the charge the electron sees as a result of the other $K$ -shell electrons. Suppose we want to detect the presence of aluminum, copper, and tungsten in a sample. a. Estimate the minimum potential difference through which the bombarding electron beam must be accelerated to produce the characteristic X-rays of each of these elements. b. From the data provided in Table 29.1 , estimate the energy of an electron in the $n=2$ state for each of these elements.
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A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A gascous sample is irradiated with UV light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write $$h v=\mathrm{IE}+\frac{1}{2} m u^{2}$$ where $v$ is the frequency of the UV light, and $m$ and $u$ are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be $5.34 \times 10^{-19} \mathrm{J}$ using a UV source of wavelength $162 \mathrm{nm}$ Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?
A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A sample is irradiated with UV light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write $$h v=I E+\frac{1}{2} m u^{2}$$ where $v$ is the frequency of the UV light, and $m$ and $u$ are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be $5.34 \times 10^{-19} \mathrm{J}$ using a UV source of wavelength 162 nm. Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?
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