Exercises — Hydrogen energy levels Eₙ = −13.6 - n² eV

WHAT we do: substitute $n$ into $E_n=-13.6/n^2$.

• $E_1=-13.6/1^2=\boxed{-13.6\text{eV}}$
• $E_2=-13.6/2^2=-13.6/4=\boxed{-3.40\text{eV}}$

WHY negative: we measured energy from the free-electron mark $E(\infty )=0$; a bound electron sits below that mark, so its energy is negative. $n=1$ is the deepest shelf.

FALSE. As $n$ grows, $1/n^2$ shrinks toward $0$, so $E_n=-13.6/n^2$ climbs toward $0$ (less negative). The electron is more loosely held on higher shelves.

Quick check: $E_1=-13.6$, $E_2=-3.40$, $E_3=-1.51$ — the numbers march up toward $0$. ✓

Ionization = drag the electron from its shelf up to the free mark $E=0$. $\Delta E=0-E_1=0-(-13.6)=\boxed{13.6\text{eV}}$ That is exactly why the number $13.6$ shows up everywhere — it's the depth of the bottom shelf.

WHAT & WHY: a photon carries the difference of two shelves (energy is conserved). $\Delta E=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=13.6\left(\frac{1}{4}-\frac{1}{9}\right)=13.6\cdot \frac{5}{36}=\boxed{1.89\text{eV}}$ Convert with $hc=1240\text{eV⋅nm}$ (one division turns eV into nm): $\lambda =\frac{1240}{1.89}\approx \boxed{656\text{nm}}(\text{red — the H-alpha Balmer line})$

$\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=13.6\cdot \frac{3}{4}=10.2\text{eV}$ $\lambda =\frac{1240}{10.2}\approx \boxed{122\text{nm}}(\text{ultraviolet})$ WHY UV: this is the biggest gap in the whole atom (into the deep basement), so the photon is very energetic → very short wavelength → beyond violet, in the UV.

The electron starts at $E_1=-13.6$ eV. After absorbing: $E_{\text{final}}=-13.6+12.09=-1.51\text{eV}$ Which shelf sits at $-1.51$? Solve $-13.6/n^2=-1.51\Rightarrow n^2=13.6/1.51=9\Rightarrow \boxed{n=3}$. WHY this works: the photon energy must exactly match a shelf-to-shelf gap, otherwise it isn't absorbed at all. $12.09$ eV is precisely the $1\to 3$ gap.

$E_2-E_1=-3.40-(-13.6)=10.2\text{eV}$ $E_3-E_2=-1.51-(-3.40)=1.89\text{eV}$ $E_4-E_3=-0.85-(-1.51)=0.66\text{eV}$ Pattern: gaps shrink fast — $10.2\to 1.89\to 0.66$. The shelves crowd together toward $E=0$ because $1/n^2$ flattens out.

How to read the figure below. The vertical axis is energy in eV, drawn true to scale — a shelf twice as deep is drawn twice as far down. Each solid white horizontal line is one allowed shelf $n=1,\dots ,5$; the dashed pink line at the top is $E=0$, the "free electron" mark. On the left, the blue numbers give each shelf's energy $E_n$ (the same values you just computed, $-13.6,-3.40,-1.51,\dots$). On the right, the yellow labels give the level number $n$. The yellow double-arrow measures the bottom gap $E_2-E_1=10.2$ eV — the tallest jump on the board. The pink double-arrow measures a top gap $E_4-E_3=0.66$ eV — barely a sliver. So the two arrows are literally the two numbers from your algebra, drawn as heights: the yellow arrow is about $15\times$ taller than the pink one, which is exactly $10.2/0.66$. What to observe: equal steps in $n$ give wildly unequal steps in energy — that unevenness is the whole story of the hydrogen spectrum.

WHY this tool: radii scale as $r_n=a_0{n}^2$ (derived from force balance + quantization in the parent).

• $r_1=a_0\cdot 1^2=0.0529\text{nm}$
• $r_3=a_0\cdot 3^2=9a_0=0.476\text{nm}$
• Factor: $r_3/r_1=9$. $\boxed{r_3=9r_1=0.476\text{nm}}$. Higher shelves are much farther out — that's why they're weakly bound.

Predict: any jump into $n=1$ crosses the huge bottom gap, so $3\to 1$ should win. Verify: $\Delta E_{4\to 2}=13.6\left(\frac{1}{4}-\frac{1}{16}\right)=13.6\cdot \frac{3}{16}=2.55\text{eV}$ $\Delta E_{3\to 1}=13.6\left(1-\frac{1}{9}\right)=13.6\cdot \frac{8}{9}=12.09\text{eV}$ $12.09>2.55$ ✓ — $\boxed{3\to 1}$ is far more energetic (a Lyman/UV line).

WHAT is $n=\infty$? The electron starts free ($E_{\infty }=0$) and drops to $n=2$. $\Delta E=13.6\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right)=13.6\cdot \frac{1}{4}=3.40\text{eV}$ $\lambda =\frac{1240}{3.40}\approx \boxed{365\text{nm}}$ WHY "limit": as $n_i\to \infty$ the gaps stop growing, so the photon energy caps at $3.40$ eV — the maximum energy (hence minimum $\lambda$) any Balmer line can reach.

WHY $Z^2$: the proton pull is $Z$ times stronger, and the energy carries the charge squared: $E_n=-13.6Z^2/n^2$. $E_1({\text{He}}^{+})=-13.6\cdot \frac{2^2}{1^2}=-13.6\cdot 4=\boxed{-54.4\text{eV}}$ Ionization energy $=|E_1|=\boxed{54.4\text{eV}}$ — four times hydrogen's, because the nucleus grips its single electron far harder.

$\Delta E=13.6Z^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=13.6\cdot 4\cdot \frac{3}{4}=40.8\text{eV}$ $\lambda =\frac{1240}{40.8}\approx \boxed{30.4\text{nm}}$ Compared to H's $122$ nm, it is exactly $4\times$ shorter ($122/4=30.5$). WHY: every energy is scaled by $Z^2=4$, so every wavelength shrinks by $4$. Same pattern, "zoomed in" by $Z^2$.

Step 1 — energy of the photon: $E=\frac{1240}{434}=2.857\text{eV}$ Step 2 — guess the series. Visible light (Balmer) means $n_f=2$. Test: $2.857=13.6\left(\frac{1}{4}-\frac{1}{n_i^2}\right)\Rightarrow \frac{1}{n_i^2}=\frac{1}{4}-\frac{2.857}{13.6}=0.25-0.210=0.040$ $n_i^2=\frac{1}{0.040}=25\Rightarrow n_i=5$ Answer: $\boxed{n=5\to n=2}$ (the H-gamma Balmer line). ✓ WHY start with $n_f=2$: only Balmer lines land in the visible band; picking the series first turns a two-unknown problem into a one-line solve.

Symbols we'll use (defined here so nothing is unearned):

• $e$ = the size of the electron's charge (also the proton's), $e=1.602\times 10^{-19}\text{C}$ (coulombs). It sets how strongly proton and electron attract.
• ${\epsilon }_0$ = the vacuum permittivity, a fixed constant of nature, ${\epsilon }_0=8.854\times 10^{-12}$ (in SI units). It appears in Coulomb's law and just fixes the strength-scale of the electric force. We bundle it into one symbol $k=\frac{1}{4\pi {\epsilon }_0}=8.99\times 10^9$ so the algebra stays clean.
• $r$ = orbit radius, $v$ = electron speed.

Step 0a — write the two forces separately. The proton attracts the electron by Coulomb's law with an inward force $F_{\text{Coulomb}}=k\frac{e^2}{r^2}\left(k=\frac{1}{4\pi {\epsilon }_0}\right).$ To keep moving in a circle the electron needs an inward Centripetal force $F_{\text{centripetal}}=\frac{m{v}^2}{r}.$ Step 0b — set them equal (this is the whole Bohr idea). The Coulomb pull is the centripetal force — there is no other inward push — so $k\frac{e^2}{r^2}=\frac{m{v}^2}{r}.$ Step 0c — cancel one $r$ and see WHY. The left side has $r^2$ in the denominator, the right side only $r$. Multiply both sides by $r$: the left loses one power ($r^2\to r$), the right's denominator vanishes entirely: $k\frac{e^2}{r}=m{v}^2.$ That single cancellation is what makes the next line clean. Step 1 — read off the kinetic energy. Kinetic energy is $\text{KE}=\frac{1}{2}m{v}^2$, and we just found $m{v}^2=k{e}^2/r$, so $\text{KE}=\frac{1}{2}k\frac{e^2}{r}.$ Step 2 — the potential energy. Referenced to a free electron at $E(\infty )=0$, the electric potential energy is $\text{PE}=-k\frac{e^2}{r}=-2\text{KE}.$ Step 3 — add them. $E=\text{KE}+\text{PE}=\text{KE}-2\text{KE}=-\text{KE}.$ So total energy is exactly minus the kinetic energy — the on-page justification. Step 4 — numbers at $n=1$. Since $E_1=-13.6$ eV and $E=-\text{KE}$, we get KE $=13.6\text{eV}$. Convert and invert $\text{KE}=\frac{1}{2}m{v}^2$: $\text{KE}=13.6\times 1.602\times 10^{-19}=2.179\times 10^{-18}\text{J}$ $v=\sqrt{\frac{2\text{KE}}{m}}=\sqrt{\frac{2(2.179\times 10^{-18})}{9.11\times 10^{-31}}}=\sqrt{4.78\times 10^{12}}\approx \boxed{2.19\times 10^6\text{m/s}}$ That's about $0.7\%$ of light speed — fast, but non-relativistic, so the Bohr picture holds.

WHY: since $E_{\text{photon}}=13.6(1/n_f^2-1/n_i^2)$ and $1/\lambda =E_{\text{photon}}/hc$: $\frac{1}{\lambda }=\frac{13.6}{hc}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ So the single constant $R_H=13.6/hc$ controls every line — see Rydberg constant and formula. $R_H=\frac{13.6\text{eV}}{1240\text{eV⋅nm}}=0.01097{\text{nm}}^{-1}=1.097\times 10^7{\text{m}}^{-1}$ Matches the measured Rydberg constant. ✓ The whole line spectrum is one number times a difference of two square-reciprocals.