Bohr atomic model and electron shells: Exercises — Hardware deep dive

This page is a workout for the parent topic. Every constant you need is here — nothing is assumed.

Level 1 — Recognition

Exercise 1.1

Which integer $n$ labels the ground state (lowest, most tightly bound orbit) of hydrogen, and what is its energy?

Recall Solution

WHAT: The ground state is the smallest allowed orbit. WHY: $E_{n} = - 13.6/ n^{2}$ is most negative (lowest) when $n$ is smallest, and the smallest allowed integer is $n = 1$ . $E_{1} = - \frac{13.6}{1 ^{2}} === - 13.6 eV == .$

Exercise 1.2

State the maximum number of electrons in shells $n = 1, 2, 3, 4$ (the K, L, M, N shells).

Recall Solution

Use capacity $= 2 n^{2}$ :

K ( $n = 1$ ): $2 (1)^{2} = 2$
L ( $n = 2$ ): $2 (2)^{2} = 8$
M ( $n = 3$ ): $2 (3)^{2} = 18$
N ( $n = 4$ ): $2 (4)^{2} = 32$

Answer: 2, 8, 18, 32.

Level 2 — Application

Exercise 2.1

Find the radius $r_{2}$ of the second Bohr orbit of hydrogen, in ångströms.

Recall Solution

WHAT: Plug $n = 2$ into $r_{n} = a_{0} n^{2}$ . WHY $n^{2}$ ? From the derivation, radius scales with the square of the shell number, so orbits spread out fast. $r_{2} = 0.529 \times 2^{2} = 0.529 \times 4 === 2.116 \overset{˚}{A} == .$

Exercise 2.2

An electron falls from $n = 4$ to $n = 2$ . Find the photon energy emitted.

Recall Solution

WHAT: Energy released = difference between the two levels. $E_{4} = - \frac{13.6}{16} = - 0.85 eV, E_{2} = - \frac{13.6}{4} = - 3.4 eV .$ WHY subtract this way: the electron drops to a more negative level, losing energy, which leaves as a photon: $Δ E = E_{4} - E_{2} = - 0.85 - (- 3.4) === 2.55 eV == .$

Exercise 2.3

Silicon has atomic number $Z = 14$ . Fill its shells from the inside out and state the number of valence electrons.

Recall Solution

WHAT: Pour 14 electrons into shells, filling the innermost first.

K takes 2 → 12 left
L takes 8 → 4 left
M takes 4 (it can hold 18, but only 4 remain)

Configuration: K(2) L(8) M(4), and $2 + 8 + 4 = 14$ ✔. Valence electrons = electrons in the outermost occupied shell (M) = 4. This is exactly what makes Si a semiconductor — see Semiconductors and the band gap.

Level 3 — Analysis

Exercise 3.1

The $n = 3 \to n = 2$ jump in hydrogen (the red H-alpha line) releases $1.89 eV$ . What wavelength of light is this, and is it visible?

Recall Solution

WHAT: Convert energy to wavelength with $λ (nm) = 1240/ E (eV)$ . WHY this tool: a photon's energy and its colour (wavelength) are two names for the same thing, tied by $E = h c / λ$ . Rearranged with $h c = 1240 eV\cdotnm$ it becomes a one-step division. $λ = \frac{1240}{1.89} \approx== 656 nm == .$ Visible light runs ~400–700 nm, so 656 nm sits at the red end — yes, visible. Look at Figure 1: this is the biggest gap inside the visible-light group of jumps.

Figure — Bohr atomic model and electron shells

Exercise 3.2

Compare the sizes of the gaps between consecutive levels: is $E_{2} - E_{1}$ bigger or smaller than $E_{3} - E_{2}$ ? Explain what this means for the spectral lines as $n$ grows.

Recall Solution

WHAT: Compute both gaps. $E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 eV,$ $E_{3} - E_{2} = - 1.51 - (- 3.4) = 1.89 eV .$ So $E_{2} - E_{1} === 10.2 eV ==$ is much bigger than $E_{3} - E_{2} = 1.89 eV$ . WHAT IT MEANS: the levels crowd together as $n$ rises (they all pile up toward $0$ ). So spectral lines for high- $n$ jumps bunch closer and closer — a converging series, visible as the squeezing ladder in Figure 2.

Exercise 3.3

How much energy is needed to ionize hydrogen from its ground state (rip the electron all the way out to $n = \infty$ )?

Recall Solution

WHAT: Ionization = climb from $n = 1$ to $n = \infty$ , where $E_{\infty} = - 13.6/ \infty^{2} = 0$ . $Δ E = E_{\infty} - E_{1} = 0 - (- 13.6) === 13.6 eV == .$ WHY $E_{\infty} = 0$ : a free electron infinitely far away with no motion has zero energy by our convention; every bound state sits below zero, so freeing it costs exactly $∣ E_{1} ∣$ .

Level 4 — Synthesis

Exercise 4.1

A hydrogen atom absorbs a $12.09 eV$ photon while in its ground state. Which level does the electron reach? (Give $n$ .)

Recall Solution

WHAT: After absorbing, its new energy is $E_{1} + 12.09 = - 13.6 + 12.09 = - 1.51 eV$ . WHY: absorbing a photon adds its energy to the electron, pushing it up. Match $- 1.51 = - 13.6/ n^{2} \Rightarrow n^{2} = 13.6/1.51 = 9 \Rightarrow== n = 3 == .$ Check the logic: only photons whose energy lands exactly on a level are absorbed — that's why atoms are picky about colours.

Exercise 4.2

Take a "hydrogen-like" ion with a single electron but a nucleus of charge $+ Z e$ (for example He $^{+}$ with $Z = 2$ ). The energy formula generalises to $E_{n} = - 13.6 Z^{2} / n^{2} eV$ . Find the ground-state energy of He $^{+}$ and the energy to ionize it.

Recall Solution

WHAT: Set $Z = 2$ , $n = 1$ . WHY $Z^{2}$ appears: a stronger nuclear charge pulls the electron in harder — the Coulomb force in the derivation carries a factor $Z$ , and it enters the energy squared. $E_{1} = - 13.6 \times \frac{2 ^{2}}{1 ^{2}} = - 13.6 \times 4 === - 54.4 eV == .$ Ionization energy $= ∣ E_{1} ∣ === 54.4 eV ==$ — four times harder to ionize than hydrogen. That $Z^{2}$ scaling is why heavier nuclei hold inner electrons ferociously.

Level 5 — Mastery

Exercise 5.1

An electron sits in the $n = 2$ shell of hydrogen. List every distinct photon energy that could be emitted as it cascades down to $n = 1$ , and state how many spectral lines a large population of such atoms would show. Then repeat starting from $n = 3$ .

Recall Solution

From $n = 2$ : the only downward jump is $2 \to 1$ . $E_{2} - E_{1} = - 3.4 - (- 13.6) === 10.2 eV == .$ That's 1 line.

From $n = 3$ : three possible paths exist, giving three distinct photons:

$3 \to 1$ : $- 1.51 - (- 13.6) === 12.09 eV ==$
$3 \to 2$ : $- 1.51 - (- 3.4) === 1.89 eV ==$
$2 \to 1$ : $- 3.4 - (- 13.6) === 10.2 eV ==$

So starting from level $n$ , the number of possible lines is "choose 2 from the levels $1.. n$ " $= (2 n)$ : for $n = 2$ that's $(2 2) ...$ wait — it is $(2 n)$ meaning $(2 2) = 1$ ? Let's be exact: the count is the number of distinct level-pairs below-or-at $n$ , i.e. $(2 n)$ . For $n = 2$ : $(2 2)$ is wrong; use $(2 n) = \frac{n ( n - 1 )}{2}$ → $n = 2$ gives $1$ , $n = 3$ gives $3$ . ✔

Sanity check with energy conservation: the $3 \to 1$ photon ( $12.09$ ) must equal the two-step sum $3 \to 2 \to 1$ : $1.89 + 10.2 = 12.09$ ✔. Energy is bookkept perfectly no matter the path.

Exercise 5.2

Two facts are handed to you: (a) copper has 1 valence electron, (b) silicon has 4. Using only Bohr-model shell reasoning, argue which is the better conductor and connect it to the Hardware chain in Conductors insulators and doping.

Recall Solution

WHAT the model says: valence electrons are the outermost, least-bound electrons — the ones a small energy can free into conduction. Copper (1 valence e⁻): a lone outer electron is loosely held and easily shared into a "sea" of free electrons → a good conductor. Silicon (4 valence e⁻): four electrons lock into rigid covalent bonds (each shared with a neighbour), so at low energy almost none are free → an insulator-ish semiconductor; only when you supply the band-gap energy (heat/light/voltage) do a few break loose. Conclusion: copper conducts far better. The chain: shell filling (this note) → valence count → band structure → band gap size → conductor vs. insulator. Bohr's simple shell picture is the first domino.

Wrap-up recall

Recall One-line answers (hide first)

Ground-state hydrogen energy ::: $- 13.6$ eV
$r_{2}$ in ångströms ::: $2.116$ Å
$n = 4 \to 2$ photon ::: $2.55$ eV
H-alpha wavelength ::: $\approx 656$ nm (red)
Ionization from ground state ::: $13.6$ eV
He $^{+}$ ground-state energy ::: $- 54.4$ eV
Silicon valence electrons ::: 4

Connections

Bohr atomic model and electron shells — the parent this drills.
Valence electrons and bonding — Exercises 2.3, 5.2.
Semiconductors and the band gap — the "why 4 valence" payoff.
Conductors insulators and doping — Exercise 5.2's chain.