Corrected exercises: atomic structure and electron configuration (202-SN1-RE)

The mass number is $A=37$ and the atomic number is $Z=17$: the nucleus contains 17 protons and $37-17=20$ neutrons. The neutral $^{37}Cl$ atom has 17 electrons. The $^{37}Cl^{-}$ ion has gained one electron: same nucleus (17 protons, 20 neutrons), but 18 electrons.

$M=34.9689\times 0.7577+36.9659\times 0.2423=26.497+8.956=35.45$ u. The tabulated value is not a whole number because it is not the mass of any one atom, but the average of the isotopic masses weighted by their natural abundances. It falls between 35 and 37, closer to 35 since $^{35}Cl$ is three times as abundant.

Let $x$ be the abundance of $^{63}Cu$; then $62.9296x+64.9278(1-x)=63.546$. Solving: $x=\dfrac{64.9278-63.546}{64.9278-62.9296}=\dfrac{1.3818}{1.9982}=0.6915$. Natural copper is therefore 69.15% $^{63}Cu$ and 30.85% $^{65}Cu$.

$M(Br)=78.9183\times 0.5069+80.9163\times 0.4931=79.90$ u. For $Br_{2}$, each atom is independently $^{79}Br$ (probability $p=0.5069$) or $^{81}Br$ ($q=0.4931$), which gives three peaks: $^{79}Br_{2}$ at 158 u with probability $p^{2}=0.257$, $^{79}Br^{81}Br$ at 160 u with $2pq=0.500$ (there are two ways to form the mixed molecule, hence the factor 2), and $^{81}Br_{2}$ at 162 u with $q^{2}=0.243$. The intensities are therefore in a ratio close to 1 : 2 : 1, with the central peak twice as tall as the other two. This is the classic signature of bromine in mass spectrometry.

$\nu=\dfrac{c}{\lambda}=\dfrac{3.00\times 10^{8}}{589\times 10^{-9}}=5.09\times 10^{14}$ Hz. The photon energy is $E=h\nu=(6.626\times 10^{-34})(5.09\times 10^{14})=3.37\times 10^{-19}$ J. For one mole: $3.37\times 10^{-19}\times 6.022\times 10^{23}=2.03\times 10^{5}$ J/mol, that is 203 kJ/mol.

At the threshold, all of the photon energy goes exactly into extracting the electron: $\dfrac{hc}{\lambda_{0}}=W_{0}$, so $\lambda_{0}=\dfrac{hc}{W_{0}}=\dfrac{(6.626\times 10^{-34})(3.00\times 10^{8})}{3.68\times 10^{-19}}=5.40\times 10^{-7}$ m, that is 540 nm: green light. Any light with a wavelength below 540 nm (hence more energetic) ejects electrons; any light beyond it ejects none.

$E=\dfrac{hc}{\lambda}=\dfrac{(6.626\times 10^{-34})(3.00\times 10^{8})}{400\times 10^{-9}}=4.97\times 10^{-19}$ J. The maximum kinetic energy is the excess: $E_{k}=E-W_{0}=4.97\times 10^{-19}-3.68\times 10^{-19}=1.29\times 10^{-19}$ J. Then $v=\sqrt{\dfrac{2E_{k}}{m_{e}}}=\sqrt{\dfrac{2(1.29\times 10^{-19})}{9.109\times 10^{-31}}}=5.32\times 10^{5}$ m/s.

At 700 nm, $E=\dfrac{hc}{\lambda}=2.84\times 10^{-19}$ J, which is below $W_{0}=3.68\times 10^{-19}$ J: no electron is ejected, and the intensity changes nothing. In the wave model, the energy carried by a wave depends on its amplitude, hence on its intensity: with light strong enough, or shone long enough, the required energy should eventually build up. Experiment says the opposite, and instantly: what counts is the energy of a single photon, $h\nu$. Raising the intensity raises the number of photons, not the energy of each one. This is exactly Einstein's argument of 1905.

$n$ (principal): the energy level and the size of the orbital; $n=1,\ 2,\ 3,...$ $l$ (azimuthal): the shape of the orbital and the subshell ($l=0$ for s, 1 for p, 2 for d, 3 for f); $l$ runs from $0$ to $n-1$. $m_{l}$ (magnetic): the orientation of the orbital, that is, the particular orbital; $m_{l}$ takes the integers from $-l$ to $+l$. $m_{s}$ (spin): the direction of the electron's spin; $m_{s}=+\tfrac{1}{2}$ or $-\tfrac{1}{2}$.

The $n=4$ level contains $n^{2}=16$ orbitals and at most $2n^{2}=32$ electrons.

The $3d$ subshell has $l=2$, hence $2l+1=5$ orbitals and at most $2(2l+1)=10$ electrons.

(i) Allowed: $l=2<n=3$, $m_{l}=-1$ lies in $[-2, 2]$, $m_{s}=+\tfrac{1}{2}$. (ii) Forbidden: $l$ must be strictly less than $n$ (at most $n-1=1$); here $l=2$ with $n=2$ is impossible. (iii) Forbidden: $m_{l}$ must lie between $-l$ and $+l$, that is, between $-3$ and $+3$; $m_{l}=-4$ is outside that interval. (iv) Forbidden: $m_{s}$ can only be $+\tfrac{1}{2}$ or $-\tfrac{1}{2}$, never $+1$.

Phosphorus ($Z=15$): $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$, that is $[Ne]3s^{2}3p^{3}$. The $Ca^{2+}$ ion: neutral calcium is $[Ar]4s^{2}$; by losing its two $4s$ electrons, the ion becomes $[Ar]$, that is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$ (18 electrons).

Phosphorus has 5 valence electrons (the $n=3$ shell: $3s^{2}3p^{3}$). By Hund's rule, the three $3p$ electrons each occupy a separate orbital with the same spin: phosphorus therefore has 3 unpaired electrons.

Chromium: $[Ar]4s^{1}3d^{5}$ (not $4s^{2}3d^{4}$); copper: $[Ar]4s^{1}3d^{10}$ (not $4s^{2}3d^{9}$). One $4s$ electron moves into $3d$ because a half-filled ($3d^{5}$) or completely filled ($3d^{10}$) $d$ subshell is especially stable.

Iron ($Z=26$): $[Ar]4s^{2}3d^{6}$. The six $3d$ electrons are spread over the five $d$ orbitals: five occupy one orbital each (Hund), the sixth pairs up, which leaves 4 unpaired electrons. Since it has unpaired electrons, iron is paramagnetic.

$E_{2}=-\dfrac{2.18\times 10^{-18}}{4}=-5.45\times 10^{-19}$ J and $E_{3}=-\dfrac{2.18\times 10^{-18}}{9}=-2.42\times 10^{-19}$ J.

The electron drops from a higher level to a lower one: the atom emits energy. $\Delta E=E_{2}-E_{3}=-5.45\times 10^{-19}-(-2.42\times 10^{-19})=-3.03\times 10^{-19}$ J. The negative sign confirms emission; the energy of the emitted photon is $|\Delta E|=3.03\times 10^{-19}$ J.

$\lambda=\dfrac{hc}{|\Delta E|}=\dfrac{(6.626\times 10^{-34})(3.00\times 10^{8})}{3.03\times 10^{-19}}=6.57\times 10^{-7}$ m, about 657 nm. This wavelength lies in the visible range (red light): it is the characteristic red line of the hydrogen spectrum. Every visible line, in fact, ends on level $n=2$, that of the Balmer series; those that land on $n=1$, the Lyman series, belong to the ultraviolet.

Across a period, the number of shells does not change but the number of protons increases: the effective nuclear charge felt by the valence electrons grows, pulls them in harder, and the radius decreases. Down a group, one shell is added on each row: the valence electrons are farther and farther from the nucleus and shielded by the inner shells, so the radius increases.

By increasing radius: $Cl<Mg<Na<K$. Chlorine is farthest to the right (maximum $Z_{eff}$, smallest); $Mg$ then $Na$ are in the same period but $Na$ is farther left and so larger; potassium has one more shell and is the largest.

The $Al<Mg$ anomaly: magnesium has the configuration $[Ne]3s^{2}$, a full and stable $3s$ subshell that is hard to ionize; aluminum $[Ne]3s^{2}3p^{1}$ easily loses its single $3p$ electron, which is higher in energy. The $S<P$ anomaly: phosphorus has $3p^{3}$, a stable half-filled subshell (one electron per orbital); sulfur $3p^{4}$ has one doubly occupied orbital, and the repulsion between those two paired electrons makes removal easier, hence a lower ionization energy.

All have 10 electrons, so it is the number of protons that sets them apart. Nuclear charges: $O$ 8, $F$ 9, $Na$ 11, $Mg$ 12. The more protons for the same cloud of 10 electrons, the more contracted the ion. By decreasing radius: $O^{2-}>F^{-}>Na^{+}>Mg^{2+}$.

$E_{1}=-\dfrac{(2.18\times 10^{-18})(2^{2})}{1^{2}}=-8.72\times 10^{-18}$ J. To ionize is to bring the electron to $n=\infty$, where $E=0$: $8.72\times 10^{-18}$ J must be supplied per ion, that is $8.72\times 10^{-18}\times 6.022\times 10^{23}=5.25\times 10^{6}$ J/mol, or 5251 kJ/mol. This is exactly 4 times the value for hydrogen, because the energy is proportional to $Z^{2}$ and $Z$ goes from 1 to 2. Physically, the helium nucleus attracts its single electron twice as strongly, and the electron is much harder to remove.

$|\Delta E|=2.18\times 10^{-18}\left(\dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}\right)=2.18\times 10^{-18}\times 0.75=1.635\times 10^{-18}$ J. Then $\lambda=\dfrac{hc}{|\Delta E|}=\dfrac{1.988\times 10^{-25}}{1.635\times 10^{-18}}=1.22\times 10^{-7}$ m, that is 122 nm: in the far ultraviolet. The whole Lyman series is invisible to the eye.

$|\Delta E|=2.18\times 10^{-18}\left(\dfrac{1}{9}-\dfrac{1}{16}\right)=2.18\times 10^{-18}\times 0.0486=1.06\times 10^{-19}$ J, so $\lambda=\dfrac{1.988\times 10^{-25}}{1.06\times 10^{-19}}=1.88\times 10^{-6}$ m, that is 1876 nm: in the infrared. By decreasing energy: Lyman, then Balmer, then Paschen. The reason is that the levels crowd together as $n$ increases: the drop to $n=1$ crosses the largest energy gap, the drop to $n=3$ the smallest. Ultraviolet, visible, infrared: the order of the series is also the order of the spectrum. Each series is named after its final level: Lyman falls to $n=1$ (ultraviolet), Balmer to $n=2$ (visible), Paschen to $n=3$ (infrared).

For $He^{+}$, $|\Delta E|=2.18\times 10^{-18}\times 2^{2}\left(\dfrac{1}{2^{2}}-\dfrac{1}{4^{2}}\right)=2.18\times 10^{-18}\times 4\times \dfrac{3}{16}=2.18\times 10^{-18}\times 0.75$, which is the same value $1.635\times 10^{-18}$ J as in b), hence the same wavelength of 122 nm. In general, $|\Delta E|=2.18\times 10^{-18}Z^{2}\left(\dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}\right)$: if $n_{f}$ and $n_{i}$ are replaced by $Zn_{f}$ and $Zn_{i}$, the factor $Z^{2}$ cancels exactly against the squares in the denominators. In other words, every line of hydrogen has a twin in $He^{+}$ (levels doubled), in $Li^{2+}$ (levels tripled), and so on. This is why some stellar lines attributed to hydrogen were at first misidentified.

$\lambda=\dfrac{h}{m_{e}v}=\dfrac{6.626\times 10^{-34}}{(9.109\times 10^{-31})(5.00\times 10^{6})}=\dfrac{6.626\times 10^{-34}}{4.554\times 10^{-24}}=1.45\times 10^{-10}$ m. This is the same order of magnitude as the diameter of an atom and as the spacing between two planes of a crystal: the electron can therefore be diffracted by a crystal, exactly like X-rays. This is the principle of electron diffraction and, further on, of the electron microscope, whose resolving power far exceeds that of the optical microscope.

$\lambda=\dfrac{h}{mv}=\dfrac{6.626\times 10^{-34}}{(0.145)(40.0)}=\dfrac{6.626\times 10^{-34}}{5.80}=1.14\times 10^{-34}$ m. This wavelength is about $10^{24}$ times smaller than that of the electron, and incomparably smaller than any object or slit one could make. No diffraction phenomenon can therefore be observed: the wave nature exists in principle for the ball, but it is experimentally out of reach. It is the mass in the denominator that crushes everything.

$\Delta v\geq \dfrac{h}{4\pi m_{e}\Delta x}=\dfrac{6.626\times 10^{-34}}{4\pi (9.109\times 10^{-31})(1.0\times 10^{-10})}=5.8\times 10^{5}$ m/s. This uncertainty is more than a tenth of the speed in part a): we do not even know the order of magnitude of the electron's speed once we know it is in the atom. For the ball: $\Delta v\geq \dfrac{6.626\times 10^{-34}}{4\pi (0.145)(1.0\times 10^{-3})}=3.6\times 10^{-31}$ m/s, a perfectly negligible uncertainty.

A Bohr orbit assumes that the position and the speed of the electron are known at every instant: it is a trajectory. But part c) shows that locating the electron within the atom makes its speed totally undetermined, and the reverse is true too. The trajectory therefore has no meaning at this scale, not for lack of instruments, but by nature. What remains computable is the probability of finding the electron at a point: it is the square of the wave function, and the region where this probability is significant is called an orbital. The Bohr model keeps its energies $E_{n}$, which are correct for hydrogen, but loses its circular orbits. For the baseball, the uncertainty is so small that the classical trajectory remains perfectly valid: quantum mechanics does not contradict Newtonian mechanics, it contains it.