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All Solutions of Type: Atomic
 0 Click here to jump to the problem! GR8677 #26 Atomic$\Rightarrow$}X-Rays The K-series refers to the inner-most shell. (The order from inner to outer goes like K, L, M, N.) K-series refers to a transition from some outer state to the inner-most shell, where $n=1$ in the usual Bohr equation, $E=-13.6(Z-1)^2 \left(1-1/n_i^2\right) eV, $ where $(Z-1)^2$ is used to account for shielding. For electrons bombarding a target, one assumes that the electrons are coming from $n_i=\infty$, thus the equation becomes $E=-13.6(Z-1)^2\approx 10*900\approx 10,000$ eV, as in choice (D). Click here to jump to the problem!
 1 Click here to jump to the problem! GR8677 #30 Atomic$\Rightarrow$}Orbitals Recall that: $\begin{eqnarray} s&=&0\\ p&=&1\\ d&=&2\\ f&=&3\\ g&=&4...(etc.), \end{eqnarray}$ where the right side of the equal sign is the angular momentum $l$. And then there's also the formula $2(2l+1)$, which determines the number of electrons in each subshell. (A) The configuration does not specify anything about $3d$. This is the first row of transitional metals. (B) From the formula above, one sees that the $s$ subshell needs $2$ electrons to be filled. The $4s$ subshell is thus not filled. (C) $l=4$ corresponds to $g$. The configuration does not specify $g$. (D) Because the problem states that potassium is in the ground state, the atomic number is the same as the number of electrons in the configuration. The sum of the superscripts is 19. (E) The valance shell is $4s^2$. Many-electron atoms have wave functions that take after the hydrogen atom, and one recalls that $\psi_{40m}$ looks like a spherically symmetrical blob of a ball. This is it. Click here to jump to the problem!
 2 Click here to jump to the problem! GR8677 #41 Atomic$\Rightarrow$}Binding Energy Nickel and Iron are the most tightly bound nuclei, thus have the highest binding energy. Nickel isn't on the list, thus Iron must be the choice. Click here to jump to the problem!
 3 Click here to jump to the problem! GR8677 #48 Atomic$\Rightarrow$}Transition Transitions go from state one to another state that's completely unique relative to the first state. If both initial and final states have spherically symmetrical wave functions, then they have the same angular momentum. This is forbidden. Click here to jump to the problem!
 4 Click here to jump to the problem! GR8677 #82 Atomic$\Rightarrow$}Emission Lines (A) The Stern-Gerlach effect has to do with splitting of beams of atoms sent through an inhomogenous magnetic field. Does not have to do with emission spectrum. (B) The Stark effect has to do with energy shifts via placing atom(s) in an Electric field. (C) Splitting is often due to electron spin magnetic moments. (D) The condition is much too rigid. (E) Splitting of energies means more lines than before. This choice is general enough to be true. (The effect of energy shifts due to placing an atom in a magnetic field is called the Zeeman Effect, which isn't listed.) Click here to jump to the problem!
 5 Click here to jump to the problem! GR8677 #83 Atomic$\Rightarrow$}Spectral Line The less dense a gas is, the more precise (thin) its emission lines. So, when one has a dense gas, one expects the spectral lines to be broader---as in choice (C). Click here to jump to the problem!
 6 Click here to jump to the problem! GR8677 #84 Atomic$\Rightarrow$}Spectroscopic Notation Given the order in which energy levels are filled in atomic configuration $1s 2s 2p 3s 3p 4s 3d$ and the number of electrons in sodium, one can fill it up like $1s^2 2s^2 2p^6 3s^1$. There is a net spin from the missing electron in the $3s$ valance shell, and thus $S=1/2$. The valance $3s$ shell has $l=0$ (and $L=S$), since $s=0;p=1;d=2;f=3$. $J=l+S=1/2$. Thus, the form should be $^{2(1/2)+1}S_{1/2}=^{2}S_{1/2}$ Not even knowing anything about spectroscopic notation, one can deduce the right answer as well as the general form: $^{2S+1}(L)_{J=l+S}$, where $L$ can be either $S,P,D,F...$ depending on whether $l=0,1,2,3...$ Click here to jump to the problem!
 7 Click here to jump to the problem! GR8677 #99 Atomic$\Rightarrow$}Positronium Interestingly, this is almost an exact repeat of Exam 9677 Prob 12. The reasoning's the same: The positronium atom involves a positron-electron combination instead of the usual proton-electron combo for the H atom. Charge remains the same, and thus one can approximate its eigenvalue by changing the mass of the Rydberg energy (recall that the ground state of the Hydrogen atom is 1 Rydberg). Recall the reduced mass $\mu=\frac{m_1 m_2}{m_1+m_2}$, where for identical masses, one obtains $\mu = m/2$. The Rydberg in the regular Hydrogen energy eigenvalue formula $E=R\left(1/n_f^2 - 1/n_i^2 \right)$ is $R\propto \mu$, where $\mu \rightarrow \mu/2$. Substitute in the new value of the reduced mass to get $E\approx R/2$. $R=-13.6$ eV, and thus $E\approx -6.8$ eV. Click here to jump to the problem!
 8 Click here to jump to the problem! GR8677 #9 Atomic$\Rightarrow$}Rydberg Energy $\frac{1}{\lambda}=R\left( \frac{1}{n_f^2} - \frac{1}{n_0^2} \right)$. Given the information that the Lyman Series is $n_f=1$, and the Balmer series is $n_f=2$, one forms the ratio $\lambda_L/\lambda_B=0.25$ (taking $n_i=\infty$). This is closest to choice (A). (Recall that ETS wants the answer that best fits.) Click here to jump to the problem!
 9 Click here to jump to the problem! GR8677 #11 Atomic$\Rightarrow$}Stern-Gerlach Recall the Stern-Gerlach experiment, where (in its original set-up) a beam of neutral silver atoms are sent through an inhomogeneous magnetic field. The beam's split into two---classically, from the Lorentz force, one wouldn't expect anything to happen since all the atoms are neutral, but if one accounts for the Larmor precession, one would expect the beam to be deflected into a smear. Instead, however, the beam deflects into $2s+1$ beams, and thus this supports the idea that electrons are of spin-1/2. (Ag has one unpaired electron in its $p$ orbital.) With a beam of hydrogen atoms, one should also get a split into two, since $s=1/2$ from the electron. Click here to jump to the problem!
 10 Click here to jump to the problem! GR8677 #12 Atomic$\Rightarrow$}Positronium The positronium atom involves a positron-electron combination instead of the usual proton-electron combo for the H atom. Charge remains the same, and thus one can approximate its eigenvalue by changing the mass of the Rydberg energy (recall that the ground state of the Hydrogen atom is 1 Rydberg). Recall the reduced mass $\mu=\frac{m_1 m_2}{m_1+m_2}$, where for identical masses, one obtains $\mu = m/2$. The Rydberg in the regular Hydrogen energy eigenvalue formula $E=R\left(1/n_f^2 - 1/n_i^2 \right)$ is proportional to $\mu$. Substitute in the new value of the reduced mass to get $E\approx R/2$. $R=-13.6$ eV, and thus $E\approx -6.8$ eV. Click here to jump to the problem!
 11 Click here to jump to the problem! GR8677 #39 Atomic$\Rightarrow$}Ionization Potential Atoms with full shells have high ionization potentials---they would hardly want to lose an electron, and thus it would take a great amount of energy to ionize them. Atoms with close-to-full-shells have similarly high potentials, as compared to atoms that are in the middle of the spectrum, take $Cs^55$, for example. (A) He has a full orbital, and thus its ionization potential must be high. (B) N has a close-to-full orbital. (C) O has a close-to-full orbital. (D) Ar is a nobel gas, and thus its orbital is full. (E) This is it. Cs has the lowest IP of all (of the above). Click here to jump to the problem!
 12 Click here to jump to the problem! GR8677 #41 Atomic$\Rightarrow$}Spectroscopic Notations Recall the selection rules $\Delta m=0,\pm 1$ and $\Delta l=\pm 1$ for the electric dipole approximation in time-dependent perturbation theory. (A) This is allowed since $\Delta l = -1$ and $\Delta m = 1$ (B) This is not allowed since $\Delta l=0$, which goes against the condition that $\Delta l=\pm 1$. (C) $l=1$ for p orbitals. Recall $(s,p,d,f)=(0,1,2,3)$ (D) An electron has $s=1/2$, thus one can't have $j=l$ or $s=3/2$. (E) One does not know this for sure. Choice (A) is the best choice. Click here to jump to the problem!
 13 Click here to jump to the problem! GR8677 #17 Atomic$\Rightarrow$}Orbital Config First, Check that all choices have superscripts that add to 15. That's apparently so, so one can't easily eliminate choices based on that. Since the problem has electrons in ground-state, choice (D) and (A) cannot be it--no electrons are promoted from the s to higher orbitals. s has to be filled first. Choice (E) skips the 3s states completely, so that's out. One should have an approximate idea of what the periodic table looks like. The transitional (d-state) elements do not occur until after 4s. Even so, the first row of transitional elements is labeled 3d. Since Phosphorous does not have enough electrons to qualify for a 4s or more state, choice (B) must be right. Click here to jump to the problem!
 14 Click here to jump to the problem! GR8677 #18 Atomic$\Rightarrow$}Bohr Theory One can calculate the ionization energy from finding the ionization energy for He for He+, then subtracting it from the given ionization energy. Since He+ is a hydrogenic atom, one can apply Bohr Theory. In Bohr Theory, the energy to remove an electron is $E=Z^2 E_1$. For Helium, since it has two protons, $Z=2$. Thus, $E=4 E_1 \approx 52$ is the ionization energy. ($E_1=13.6eV$ is just 1 Rydberg or the energy of the ground state of Hydrogen.) Subtract this from the ionization energy for He given in the problem to be 79 eV, to get an answer close to choice (A). Click here to jump to the problem!
 15 Click here to jump to the problem! GR8677 #21 Atomic$\Rightarrow$}Bohr Formula One applies the Bohr formula $E\propto (1/n_f^2 - 1/n_i^2) \propto 1/\lambda$. For the Lyman radiation, this is $E_L \propto (1-1/4)$. For the Balmer radiation, this is $E_B \propto (1/4-1/9)$. Take the ratio of that to get $E_L/E_B = \frac{3/4}{5/36}=27/5$. But, since the problem wants the ratio of the wavelengths, which has an inverse relation to energy, one takes the inverse of that to get choice (B). Click here to jump to the problem!
 16 Click here to jump to the problem! GR8677 #30 Atomic$\Rightarrow$}Hydrogen Atom The hydrogen atom wave-functions all have to do with an exponentially decreasing radial function. Thus, choose choice A. (FYI: Only the spherical harmonic angular parts have trig involved, and thus choice.) Click here to jump to the problem!
 17 Click here to jump to the problem! GR8677 #31 Atomic$\Rightarrow$}Positronium The positronium atom ground-state energy can be deduced from recalling the reduced mass. $\mu=m/2$, since one has a positron and electron orbiting each other, and thus the energy, which is related to mass, is half of that of Hydrogen. $E_{positronium}=-13.6/2 eV = -6.8eV$ The Bohr formula still applies, and thus $E=E_1 \left(1/n_f^2 - 1/n_i^2 \right)=8E_1/9$, which is choice (A). Click here to jump to the problem!
 18 Click here to jump to the problem! GR8677 #64 Atomic$\Rightarrow$}Radiation Spectra Choice-by-choice analysis gives... (A) The word nuclear sounds questionable, as lines are often due to just spin-splitting. (B) The wavelengths of absorption spectra are often close to that of emissions spectrum (or overlapping). This is true. (C) Absorption spectra is often used in astrophysics to determine the contents of stars. Even though mere mostly harmless Earthlings have never visited the sun, the absorption spectra of the sun helps determine its component elements. (D) Same idea as in the previous choice. (E) A single atom does not have band spectra. So, yes, band-spectra are due to molecules. (To wit: it only has, say, 2s, 3p, etc., states, and not states that are split in-between---molecules have more degrees of freedom, i.e., they can rotate and vibrate, while atoms can't.) The remaining choice is (A). Take that. Click here to jump to the problem!
 19 Click here to jump to the problem! GR8677 #2 Atomic$\Rightarrow$}Bragg Diffraction Recall the Bragg Diffraction dispersion relation, $\lambda = 2d\sin\theta, $ thus the maximal wavelength $\lambda$ would be $2d$, choice (D). (One can derive that even if one does not remember the formula. Consider two lattice planes. View them from the side so that they appear as two parallel lines. A wave would hit the both planes at, say, an angle $\theta$ from the normal. The wave that reflects off the bottom lattice will have to travel an extra distance, relative to the wave hitting the top plane, equal to $2d\sin\theta$.) Click here to jump to the problem!
 20 Click here to jump to the problem! GR8677 #24 Atomic$\Rightarrow$}Bonding Solid Argon is a Nobel gas. It has a full shell of outer electrons, and thus it cannot bond in anything but van der Walls bonding, which isn't really bonding, but more weak like charge-attraction. One can arrive at this choice by elimination: (A) Ionic bonding occurs when one atom is a positive ion and the other the compensating negative ion. Since solid Argon isn't an ion, it can't do this. (B) Covalent bonding occurs when electrons are shared between atoms. This only happens when the atom has unfilled orbitals. (Incidentally, it only occurs when two electrons are of opposite spins due to the Pauli Exclusion Principle. That is, they must have different quantum numbers so that they can both remain stable in a low energy state.) (C) No partial charge-analysis needed. (D) Argon isn't a metal. (E) This is the one that remains. Click here to jump to the problem!
 21 Click here to jump to the problem! GR8677 #31 Atomic$\Rightarrow$}Spectroscopic Notations Spectroscopic notation is given by $^{2s+1}L_j$, and it's actually quite useful when one is dealing with multiple particles. $L\in (S,P,D,F)$, respectively, for orbital angular momentum values of $0,1,2,3$. $s=1/2$ for electrons. j is the total angular momentum. Knowing the convention, one can plug in numbers to solve $3=2s+1 \Rightarrow s=1$. Since the main-script is a S, $l=0$. The total angular momentum is $j=s+l=1$. Click here to jump to the problem!
 22 Click here to jump to the problem! GR8677 #58 Atomic$\Rightarrow$}Orbitals Eliminate (E) immediately since the superscripts do not add to 11. Each superscript stands for an electron. Eliminate (B) because the s orbital can only carry 2 electrons. Ground state means none of the electrons are promoted, and there are no states with unfilled gaps in them. Eliminate (A) since it promotes the 2p electron to 3s, leaving a unfilled orbital of lower energy. Eliminate (D) since it promotes the 3s electron to 3p, leaving an empty orbital of lower energy. Choice (C) is it. Click here to jump to the problem!
 23 Click here to jump to the problem! GR8677 #59 Atomic$\Rightarrow$}Orbital The ground state of Helium has $1s^2$ which is $l=0$ $n=1$. However, because both electrons are in the same l and n state, the Pauli Exclusion Principle (no two electrons can have exactly the same quantum number) requires that one have $s=1/2$ and the other has $s=-1/2$ for a combined total spin of $s=0$, as in a spin singlet. Thanks to user cakedamber for pointing this out. (Compare with things in the p orbitals, which have $l=1$, allowing for $m_l=-1,0,1$.) Click here to jump to the problem!
 24 Click here to jump to the problem! GR8677 #76 Atomic$\Rightarrow$}Orbitals The total angular momentum is given by $j=l+s$ where l is the orbital angular momentum and s is the spin angular momentum. (Note that, to an extent, l and s can be viewed as magnitudes, while $m_l$ and $m_s$ as directions.) The total orbital angular momentum is just $0+1+1$, since one should recall that $(s,p,d,f) \in (0,1,2,3)$. The spin angular momentum is just $1/2+1/2+1/2$ because one has three electrons.(Electrons are fermions that have spin $1/2$.) Thus, the total angular momentum is $j=2+3/2=7/2$, as in choice (A). Click here to jump to the problem!

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