"Atomic Structure" — Practice

Question 1

Calculate the wavelength of the first line of the Lyman series in hydrogen.

Accepted Answer

The first Lyman line is $n = 2 	o 1$. From the Rydberg formula, $1/\lambda = 1.097	imes10^{7}(1 - 1/4) = 8.228	imes10^{6}$ m$^{-1}$, so $\lambda = 1.215	imes10^{-7}$ m, that is 121.5 nm, in the ultraviolet.

Question 2

How many spectral lines are observed when hydrogen atoms de-excite from the $n = 6$ level by all available routes?

Accepted Answer

The count is $n(n-1)/2 = 6	imes5/2 = 15$ lines. This is the number of ways to choose two of the six levels, since every downward pair gives one line. Note the question asks for distinct lines, not transitions per atom.

Question 3

Find the radius and energy of the $n = 3$ orbit of He⁺.

Accepted Answer

Helium has $Z = 2$ and He⁺ has one electron, so Bohr applies exactly. The radius is $0.529	imes9/2 = 2.38$ Å and the energy is $-13.6	imes4/9 = -6.04$ eV. Compare hydrogen's $n = 3$ level at $-1.51$ eV: four times deeper for the same quantum number.

Question 4

The ionisation energy of hydrogen is 13.6 eV. What is the ionisation energy of an electron in the $n = 2$ level, and what wavelength would just achieve it?

Accepted Answer

The energy at $n = 2$ is $-13.6/4 = -3.4$ eV, so 3.4 eV is needed. Using $hc = 1240$ eV nm, the wavelength is 365 nm. This is the Balmer series limit, and light of shorter wavelength ionises the atom and produces a continuous spectrum.

Question 5

Calculate the de Broglie wavelength of an electron accelerated through 400 V.

Accepted Answer

Use $\lambda = 12.27/\sqrt{V}$ Å. The square root of 400 is 20, so $\lambda = 0.614$ Å. This is smaller than typical atomic spacings, which is why higher accelerating voltages give electron microscopes better resolution.

Question 6

A cricket ball of mass 150 g moves at 40 m s$^{-1}$. Find its de Broglie wavelength and comment.

Accepted Answer

The momentum is $0.150	imes40 = 6.0$ kg m s$^{-1}$, so $\lambda = 6.626	imes10^{-34}/6.0 = 1.1	imes10^{-34}$ m. That is roughly twenty orders of magnitude smaller than a nucleus, so no aperture could ever diffract it. Wave behaviour is not absent for large objects, merely undetectable.

Question 7

Explain, using the uncertainty principle, why the Bohr model cannot be correct even though it predicts hydrogen's spectrum exactly.

Accepted Answer

A Bohr orbit specifies both radius and speed exactly, so $\Delta x$ and $\Delta p$ are both zero and their product is zero, while the principle requires at least $h/4\pi$. Getting the right answer for the wrong reason is possible: Bohr's success came from the quantisation condition, which survives in quantum mechanics, rather than from the orbit, which does not.

Question 8

How many radial, angular and total nodes does a 5f orbital have?

Accepted Answer

For 5f, $n = 5$ and $l = 3$. Angular nodes equal $l$, which is 3. Radial nodes are $n-l-1 = 1$. Total nodes are $n-1 = 4$, which agrees with $3+1$.

Question 9

Which of these sets of quantum numbers is not allowed, and why: (a) $n=3$, $l=2$, $m=-2$; (b) $n=2$, $l=2$, $m=0$; (c) $n=4$, $l=0$, $m=0$?

Accepted Answer

Set (b) is not allowed. The azimuthal quantum number may only run from 0 to $n-1$, so for $n=2$ the largest permitted $l$ is 1 and a 2d orbital does not exist. Sets (a) and (c) describe a 3d and a 4s orbital respectively, and in each case $m$ lies within $-l$ to $+l$.

Question 10

Explain why 4s fills before 3d despite having a higher principal quantum number.

Accepted Answer

Orbital energy in a multi-electron atom depends on $n+l$, not on $n$ alone. For 4s, $n+l = 4$; for 3d it is 5, and the lower value fills first. Physically, the 4s orbital penetrates closer to the nucleus and is screened less effectively, which lowers its energy below 3d. In hydrogen, with no other electrons to screen, the two are not ordered this way at all.

Question 11

Write the ground-state configurations of Cr, Cu and Fe²⁺, and state the number of unpaired electrons in each.

Accepted Answer

Chromium is $[\mathrm{Ar}]3d^{5}4s^{1}$ with six unpaired electrons. Copper is $[\mathrm{Ar}]3d^{10}4s^{1}$ with one. Iron is $[\mathrm{Ar}]3d^{6}4s^{2}$, and Fe²⁺ loses the two 4s electrons to give $[\mathrm{Ar}]3d^{6}$ with four unpaired. Removing from 4s rather than 3d is the step most often got wrong.

Question 12

A metal has a work function of 2.0 eV. Will light of wavelength 700 nm eject electrons from it? Justify with a calculation.

Accepted Answer

The photon energy is $1240/700 = 1.77$ eV, below the work function of 2.0 eV, so no electrons are ejected however intense the light or however long it is applied. The threshold wavelength is $1240/2.0 = 620$ nm, so only light shorter than 620 nm will work.

Practice — "Atomic Structure"

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