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Chapter 8: Problem 57

The \(K_{a}\) line in copper is a very common one to use in \(X\) -ray ciystallography. To produce it, electrons are accelerated through a potential difference and smashed into a copper target. Section 7.8 gives the energies in a hydrogenlike atom as \(Z^{2}\left(-13.6 \mathrm{eV} / \mathrm{n}^{2}\right)\). Making the reasonable approximation that an \(n=1\) electron in copper orbils the nucleus and half of its fellow \(n=1\) electron. being unaff ected by the roughly spherical cloud of other electrons around it, estimate the minimum accelerating potential needed to make a hole in copper's \(K\) shell.

Short Answer

Expert verified
The minimum accelerating potential needed to make a hole in copper's K shell is approximately \(119.34 V\).

Step by step solution

01

Find the Energy of the K Shell

Using the formula given with Z=29/2 and n=1 (the K shell, based on the Bohr's model), the energy (E) can be calculated as follows: \(E = -Z^{2}(13.6eV/(n^{2})) = -(29/2)^{2}(-13.6eV/(1^{2})) = -119.34eV.\) The negative sign indicates the bound state of the electron.
02

Find the Accelerating Potential

The energy E is given by \(E = qV\), where q is the charge of the electron and V is voltage. Therefore, the accelerating potential \(V = E/q\). Note: The charge on an electron, \(q = 1.6 \times 10^{-19} C\). This gives \(V = -119.34 eV \times 1.6 \times 10^{-19} C = -1.91 \times 10^{-17} C V\), again the negative sign indicates the bound state of the electron.
03

Convert units from Coulomb Volt to Volt

Express your answer in volts, which is the standard unit of electric potential. The conversion factor is \(1 eV = 1.6 \times 10^{-19} C\). So, divide our result by the conversion factor to get the answer in volts: \(-1.91 \times 10^{-17} C V / (1.6 \times 10^{-19} C) = 119.34 V.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Acceleration in X-ray Crystallography
Understanding electron acceleration is central to grasping the fundamentals of X-ray crystallography. When electrons are propelled through a potential difference, they gain kinetic energy that is proportional to the voltage of the potential applied. This process is crucial in generating X-rays for crystallography.

Electrons are accelerated towards a target material, such as copper, and upon collision, their sudden deceleration leads to the emission of X-rays. The potential difference must be high enough so that when these electrons hit the target, they can knock inner shell electrons out of their orbit, creating vacancies that result in the emission of X-ray photons as other electrons drop into these lower energy states to fill the gaps.

In simple terms, imagine pulling back a slingshot - the further you pull, the more potential energy you have, and when you let go, that potential energy turns into kinetic energy as the stone flies. Similarly, a high voltage acts like a stronger pull, giving the electrons more energy to cause the necessary collisions for X-rays to be produced.
Copper K Shell and Its Role in X-ray Production
The Copper K shell refers to the innermost electron shell of a copper atom, according to the Bohr model. When we talk about making a 'hole' in the K shell, we mean ejecting one of these tightly held inner electrons, which requires a substantial amount of energy.

The energy levels of electrons in different shells are described by quantum mechanics, and in the context of copper for X-ray generation, when an electron from a higher energy level falls into the K shell, it emits an X-ray photon with energy characteristic of the copper atom - this is the basis of the method called X-ray fluorescence.

The exercise provided uses the hydrogen-like energy formula for approximation, because even in multi-electron atoms like copper, the innermost electrons feel an effective nuclear charge very similar to the actual atomic number, allowing us to treat them like hydrogen for this calculation. This simplification is the key for estimating the minimum accelerating potential needed to disrupt the K shell and generate X-ray photons for crystallography.
Bohr's Model and Its Application in the Exercise
Niels Bohr's model gives us a way to visualize an atom with discrete energy levels, which is particularly helpful when attempting to understand phenomena like X-ray generation at a basic level.

According to Bohr's model, electrons move in fixed orbits around the nucleus and can jump between these orbits by absorbing or emitting energy. The energy of an electron in a particular orbit can be calculated using the formula provided in the exercise. For copper, with an effective nuclear charge of half its atomic number due to shielding effects, we plug in the values into the Bohr's model formula to get the energy of an electron in the K shell.

This understanding allows for the calculation of the exact amount of energy required to accelerate an electron enough to knock another electron out of the K shell, a necessary step in producing the X-rays used in crystallography. The calculated energy directly correlates to the minimum voltage needed in the equipment used for this purpose, demonstrating the practical application of Bohr's theoretical model in modern scientific techniques.

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Most popular questions from this chapter

The Zeeman effect occurs in sodium just as in hydrogen - sodium's lone 3 s valence electron behaves much as hydrogen's 1.5. Suppose sodium atoms are immersed in a \(0.1 \mathrm{~T}\) magnetic field. (a) Into how many levels is the \(3 p_{1 / 2}\) level split? (b) Determine the energy spacing between these states. (c) Into how many lines is the \(3 p_{1 / 2}\) to \(3 s_{1 / 2}\) spectral line split by the field? (d) Describe quantitatively the spacing of these lines. (e) The sodium doublet \((589.0 \mathrm{nm}\) and \(589.6 \mathrm{nm}\) ) is two spectral lines. \(3 p_{3 n} \rightarrow 3 s_{1 / 2}\) and \(3 p_{1 / 2} \rightarrow 3 s_{1 / 2}\) which are split according to the two different possible spin-orbit ener gies in the \(3 p\) state (see Exercise 60 ). Detemine the splitting of the sodium doublet (the energy diff erence between the two photons). How does it compare with the line splitting of part (d), and why?

Exercise 44 gives an antisymmetric multiparticle state for two particles in a box with opposite spins. Another antisymmetric state with spins opposite and the same quantum numbers is $$ \psi_{n}\left(x_{1}\right) \downarrow \psi_{n}\left(x_{2}\right) \uparrow-\psi_{n}\left(x_{1}\right) \uparrow \psi_{n}\left(x_{2}\right) \downarrow $$ Refer to these states as \(\mid\) and 11 . We have tended to characterize exchange symmetry as to whether the state's sign changes when we swap particle labels. but we could achieve the same result by instead swapping the particles' stares, specifically the \(n\) and \(n^{\prime}\) in equation \((8-22)\). In this exercise. we look at swapping only parts of the state-spatial or spin. (a) What is the exchange symmeiry - symmetric (unchanged). antisymmetric (switching sign), or neither-of multiparticle states 1 and \(\mathrm{II}\) with respect to swapping spatial states alone? (b) Answer the same question. but with respect to swapping spin states/arrows alone. (c) Show that the algebraic sum of states 1 and \(\mathrm{II}\) may be written \(\left(\psi_{n}\left(x_{1}\right) \psi_{n}\left(x_{2}\right)-\psi_{n}\left(x_{1}\right) \psi_{n}\left(x_{2}\right)\right)(\downarrow T+\uparrow \downarrow)\) where the left arrow in any couple represents the spin of particle 1 and the right arrow that of particle 2 (d) Answer the same questions as in parts \((a)\) and (b). but for this algebraic sum. (e) Is the sum of states I and 11 still antis ymmetric if we swap the particles? total-spatial plus spin -states? (f) If the two particles repel each other, would any of the three multiparticle states - l. II, and the sum - be preferred? Explain.

In its ground state, carbon's \(2 p\) electrons interact to pro. duce \(j_{T}=0 .\) Given Hund's rule. what does this say about the total orbital angular momentum of these electrons?

Slater Determinant: A convenient and compact way of expressing multiparticle states of antisymmetric character for many fermions is the Slater determinant. lt is based on the fact that for \(N\) fermions there must be \(N\) different individual-particle states, or sets of quantum numbers. The ith state has sparial quantum numbers (which might be \(n_{i}, \ell_{i},\) and \(m_{c i}\) ) represented simply by \(n_{t}\) and spin quanturn number \(m_{s i^{i}}\). Were it occupied by the ith particle, the state would be \(\psi_{n}\left(x_{j}\right) m_{s i}\). A column corresponds to a given state and a row to a given particle. For instance, the first column corresponds to individualparticle state \(\psi_{n}(x,) m_{3},\) where \(j\) progresses (through the rows) from particle 1 to particle \(N\). The first row corresponds to particle I. which successively occupies all individual-particle states (progressing through the columns). (a) What property of determinants ensures that the multiparticle state is 0 if any two individualparticle states are identical? (b) What property of deterninants ensures that switching the labels on any two particles switches the sign of the multiparticle state?

The well-known sodium doublet is two yellow spectral lines of very close wavelength. \(589.0 \mathrm{nm}\) and \(589.6 \mathrm{nm} .\) lt is caused by splitting of the \(3 p\) energy level. due to the spin-orbit interaction. In its ground state, sodium's single valence electron is in the \(3 s\) level. It may be excited to the next higher level, the \(3 p\), then emit a photon as it drops back to the \(3 s\). However. the \(3 \rho\) is actually two levels. in which \(L\) and \(S\) are aligned and antialigned. IIn the notation of Section 8.7 these are. respectively. the \(3 p_{3 / 2}\) and the \(3 p_{1 n}\) ) Because the transitions stan from slightly different initial energies yet have identical final energies(the \(3 s\) having no orbital angular momentum to lead to spin- orbit interaction), there are two differenl wavelengths possible for the emitted photon. Calculate the difference in energy between the two photons. From this, obtain a rough value of the average strength of the internal magnetic field experienced by sodium's valence electron.
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