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 2 Click here to jump to the problem! physics_guy2018-08-08 15:16:41 The problem is just asking for the binding energy of one of the n=1 electrons in nickel, since if this electron is removed via bombardment of fast electrons, an n=2 electron will fall into its place, releasing a photon in the x-ray range. This phenomenon is known as the Auger effect, and it is used in Auger Electron Spectroscopy (AES). The photon released during this transition is known as the K-alpha line in x-ray spectroscopy. This transition was famously studied by Moseley to provide strong evidence that atomic number is identical to nuclear charge. While it is perfectly reasonable to use Moseley\'s formula for the transition from n=2 to n=1, it is not really correct since that would give you the energy of the released photon. What we really want is the binding energy of one of the two n=1 electrons, 13.6ev * Z^2. While it is also reasonable to account for the screening of nuclear charge by the other n=1 electron, the correct way to account for screening by electrons in the same orbital is to subtract 1/2 from Z, not 1. Fortunately, most of the solutions above all work out to the same order of magnitude. However, the setting of initial n to infinity shows a misunderstanding of the role of the bombarding electrons in the Auger effect, which do not undergo transitions themselves, but merely knock out core electrons with energy less than or equal to the kinetic energy of the bombarding electrons, leaving holes to be filled by higher-orbital electrons. The K-series corresponds to the removal of an n=1 electron, causing a transition from n=2 to n=1 to fill the hole. This is analogous to the Lyman-alpha transition in hydrogen.Click here to jump to the problem!
 6 Click here to jump to the problem! JasonHupp2018-07-30 11:09:59 Before the energy conversion is a hectic task to me but when i have seen here this simple tutorial on how to solve i was surprised. As i have been looking for a help and thinking is assignmenthelps.com.au legit good to help me out in this problem solving. Mean while got yours.Click here to jump to the problem!
 8 Click here to jump to the problem! jfcdu2018-07-15 14:43:27 @dberger8 the idea is that the driving frequency is set by the problem. If we want to maximize the amplitude of the current, we need to pick C such that the resonant frequency *is* the driving frequency. Thus we can set $\\omega_d = \\omega_0 = \\frac{1}{\\sqrt{LC}}$ and solve for C.Click here to jump to the problem!
 9 Click here to jump to the problem! jfcdu2018-07-15 14:40:57 @dberger8\\\\r\\\\n the idea, is that the driving frequency is set by the problem. If we want to maximize the amplitude of the current, we need to pick C such that the resonant frequency *is* the driving frequency. Thus we can set $\\\\\\\\omega_d = \\\\\\\\omega_0 = \\\\\\\\frac{1}{\\\\\\\\sqrt{LC}}$ and solve for C.Click here to jump to the problem!
 12 Click here to jump to the problem! shka2018-07-08 19:07:34 carle257 is using symmetry instead of explicitly writing out the equations for $p_y$ and $p_x$ conservation. His answer is fine and consistent with Yosun\'s, and the fact that Quark is confused by the notation $v_x = v^\\prime cos\\theta$ makes me think he doesn\'t understand how right triangles work. You can ALWAYS solve for the components of the velocity, by definition of linear momenta $p_x = mv_x$ and $p_y = mv_y$, and since $p_x$ and $p_y$ are conserved separately. Whether or not you choose to fuss with the angles is completely up to you. Since the angles and masses are identical, if you paid attention in PHYS 100 you can literally just read off the answer. Thanks Quark, for obfuscating a perfectly good solution with your utter misunderstanding of first principles. Maybe you should crack open Giancoli a few more times before applying to CERN, since I\'m pretty sure they like to conserve momentum linear momentum there.Click here to jump to the problem!
 13 Click here to jump to the problem! shka2018-07-06 00:08:13 The fraction of rays/particles detected is proportional to the solid angle subtended by the detector, so the ratio of the fractions of particles detected is equal to the ratios of the solid angles: F2 / F1 = S2 / S1. When the source is next to the detector, the solid angle is 2*Pi, since the solid angle of a whole sphere is 4*Pi, and half the rays are detected (F1 = 1/2). The formula for solid angle S is S = A / r^2 , where A is the area of the patch of sphere of radius r subtended by S. To find S2, let r = 1 meter. Since 1 meter is much larger than the radius of the detector face ( d/2 = 4 cm ), the patch of sphere subtended by S2 is almost flat, so we can use A = Pi * (d/2)^2 = 16 Pi cm^2. Thus S2 = 16 * Pi * 10^-4. Thus S2 / S1 = 8 * 10^-4. Multiplying this by F1 gives F2 = 4 * 10^-4, choice C.Click here to jump to the problem!
 14 Click here to jump to the problem! shka2018-07-03 15:22:19 fizix has asserted the converse of the spectral theorem, which is not true in general. In fact, it\'s easy to find a non-symmetric diagonalizable real matrix. For example,\r\n\r\n[ 5 -4 ] \r\n[ 4 -5 ] \r\n\r\nis ANTI-symmetric but has eigenvalues 3 and -3. The spectral theorem guarantees the diagonalizability of real symmetric matrices, but it does not force all diagonalizable real matrices to be symmetric.Click here to jump to the problem!