USAPhO/Modern Physics

The Bohr Model

Quantized energy levels, and the emission and absorption spectra they produce.

Sometimes on exams35 min read30 problems

A fingerprint made of light

Heat a tube of hydrogen gas until it glows, send the light through a prism, and you do not get a rainbow. You get four sharp coloured lines on a black background (red, blue-green, violet, deep violet) and nothing in between. Heat sodium instead and you get a different set of lines. Mercury, a third set. Every element has its own pattern, always the same, whether the gas is in a lab in Zurich or in a star ten thousand light years away.

That is a very strange fact. A hot filament gives off a smooth continuous spectrum. A hot gas of single atoms gives off a barcode. Something inside the atom can only give up energy in certain exact amounts, and those amounts are fixed properties of the element. The previous lesson on the photoelectric effect left you with a photon of energy $E = hf$ as the unit in which light is emitted and absorbed. If the atom can only emit photons of particular energies, the atom itself must only have particular energies. This lesson works out what those energies are.

Why the classical atom cannot work

Start with the picture you would draw without knowing any of this. A proton of charge $+e$ sits at the centre. An electron of mass $m$ and charge $-e$ circles it at radius $r$ and speed $v$ . The only force is Coulomb attraction, and it points inward, so it is the centripetal force:

\frac{ke^2}{r^2} = \frac{mv^2}{r}

Here $k = 8.99 \times 10^9$ N·m²/C² is the Coulomb constant. Multiply both sides by $r$ and you get $ke^2/r = mv^2$ , so the kinetic energy is

K = \tfrac12 mv^2 = \frac{ke^2}{2r}

The potential energy of two opposite charges a distance $r$ apart is $U = -ke^2/r$ , negative because you would have to do work to pull them apart. So the total energy is

E = K + U = \frac{ke^2}{2r} - \frac{ke^2}{r} = -\frac{ke^2}{2r}

Notice what fell out: $K = -U/2$ , and $E = U/2 = -K$ . That relation holds for any inverse-square orbit: it is the same one you met for satellites in gravitation, and it is called the virial theorem. It will save you arithmetic later. If a problem hands you the total energy of a bound orbit, the potential energy is twice it and the kinetic energy is minus it.

So far nothing forbids any value of $r$ . Shrink the orbit and $E$ drops; every energy from $0$ down to $-\infty$ is available, so the atom should emit a continuous spectrum. Worse, an accelerating charge radiates. A classical electron would spiral into the proton in about $10^{-11}$ seconds. Classical physics does not predict the wrong spectrum for hydrogen. It predicts that hydrogen cannot exist.

Bohr's one extra rule

In 1913 Bohr added a single restriction and changed nothing else. The electron's orbital angular momentum, $L = mvr$ , cannot take any value. It comes in whole multiples of $\hbar = h/2\pi = 1.055 \times 10^{-34}$ J·s:

L = mvr = n\hbar, \qquad n = 1, 2, 3, \ldots

The integer $n$ is the principal quantum number. Bohr had no justification for this in 1913; the reason arrives in the next lesson. Take it as a rule and follow the consequences.

You now have two equations, $ke^2/r = mv^2$ and $mvr = n\hbar$ , and two unknowns, $r$ and $v$ . Solve the second for $v = n\hbar/(mr)$ and substitute into the first:

\frac{ke^2}{r} = m \cdot \frac{n^2\hbar^2}{m^2r^2} = \frac{n^2\hbar^2}{mr^2}

Multiply through by $r^2$ and divide by $ke^2$ :

r_n = \frac{n^2\hbar^2}{mke^2} = n^2 a_0, \qquad a_0 = 0.0529\ \text{nm}

The constant $a_0$ is the Bohr radius, the size of the smallest allowed orbit. The radii go as $n^2$ : 0.0529 nm, 0.212 nm, 0.476 nm, 0.847 nm. The atom has a definite size, and that size comes out right: hydrogen really is about a tenth of a nanometre across.

Put $r_n$ back into $v = n\hbar/(mr)$ and the $n^2$ beats the $n$ :

v_n = \frac{ke^2}{n\hbar} = \frac{\alpha c}{n}, \qquad \alpha = \frac{1}{137}

The dimensionless number $\alpha = ke^2/(\hbar c)$ is the fine-structure constant. In the ground state the electron moves at $c/137 = 2.19 \times 10^6$ m/s, fast but only 0.7% of light speed, which is why ignoring relativity works as well as it does.

Figure 1. The allowed orbits, drawn to scale with each other. Drag the quantum number and watch the radius grow as n squared while the electron slows down and the orbit period stretches. Try raising Z to 2 and notice the whole atom shrinks by half.

Worth remembering

Three scalings carry most of the arithmetic in this topic: radius goes as $n^2$ , speed goes as $1/n$ , and energy goes as $1/n^2$ . The orbital period, being $2\pi r/v$ , therefore goes as $n^3$ .

The energy ladder

Substitute $r_n = n^2a_0$ into $E = -ke^2/(2r)$ :

E_n = -\frac{ke^2}{2a_0}\cdot\frac{1}{n^2} = -\frac{13.6\ \text{eV}}{n^2}

The 13.6 eV is the Rydberg energy. The minus sign means bound: you must supply energy to get the electron out to $r = \infty$ , where $E = 0$ . So the ladder reads $-13.6$ eV, $-3.40$ eV, $-1.51$ eV, $-0.850$ eV, $-0.544$ eV, and so on, crowding up towards zero. The gaps shrink fast; above $n \approx 10$ the levels are almost a continuum.

The first prediction is free. To ionise a ground-state hydrogen atom you must lift it from $-13.6$ eV to $0$ , so the ionisation energy is 13.6 eV. That was a measured number in 1913 and the model reproduced it exactly.

For a hydrogen-like ion (one electron, nuclear charge $+Ze$ ) replace $e^2$ by $Ze^2$ everywhere in the derivation. The radius picks up a $1/Z$ and the energy picks up a $Z^2$ :

r_n = \frac{n^2 a_0}{Z}, \qquad E_n = -\frac{13.6\,Z^2}{n^2}\ \text{eV}

So He⁺ ( $Z = 2$ ) has a ground state at $-54.4$ eV, and Li²⁺ ( $Z = 3$ ) sits at $-122.4$ eV in the ground state and $-13.6 \times 9/4 = -30.6$ eV at $n = 2$ . Neutral helium, with two electrons, is not covered: the electrons repel each other and the derivation collapses.

Quick check

A ground-state hydrogen atom absorbs a photon of exactly 12.75 eV. Which level does it reach?

Pick one option first.

Lines are differences

A spectrometer never sees a level. It sees a photon, and a photon appears when the electron drops from an upper level $n_i$ to a lower level $n_f$ , carrying away exactly the gap:

hf = \frac{hc}{\lambda} = E_{n_i} - E_{n_f} = 13.6\,Z^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\ \text{eV}

Work in electronvolts and nanometres and you never need $h$ or $c$ separately, because $hc = 1240$ eV·nm. Then $\lambda = 1240/\Delta E$ with $\Delta E$ in eV and $\lambda$ in nm. Memorise that one number and most of these problems become two lines of arithmetic.

Group the transitions by where they land. All the drops to $n_f = 1$ form the Lyman series, in the ultraviolet. Drops to $n_f = 2$ form the Balmer series, which is the one in the visible: those four coloured lines. Drops to $n_f = 3$ are the Paschen series, in the infrared. Within a series, the longest wavelength comes from the smallest gap ( $n_i = n_f + 1$ ) and the shortest, called the series limit, comes from $n_i = \infty$ .

Absorption is the same arithmetic backwards. Cold hydrogen sitting in front of a white-light source swallows a photon only if its energy matches a gap exactly, so the transmitted light has dark lines in precisely the places where hot hydrogen glows. That is how you read the composition of a star.

Figure 2. The energy ladder with one transition marked. Pick the upper and lower levels and read the photon energy and wavelength off the note above the figure. Try holding the lower level at 2 and walking the upper level up to 10 to watch the Balmer lines pile up against the series limit.

Worked example 1: the red line

Hydrogen drops from $n = 3$ to $n = 2$ . Find the wavelength.

Write both levels first. $E_3 = -13.6/9 = -1.511$ eV. $E_2 = -13.6/4 = -3.400$ eV. The photon carries the difference:

\Delta E = -1.511 - (-3.400) = 1.889\ \text{eV}

Then $\lambda = 1240/1.889 = 656$ nm. That is deep red, and it is the brightest line in a hydrogen tube. It has a name, H-alpha, and you will meet it again.

Worked example 2: do Balmer and Paschen overlap?

The Balmer series runs from its limit at 365 nm up to H-alpha at 656 nm. Does the Paschen series reach down into it?

The shortest Paschen wavelength is the largest Paschen gap, which is $n_i = \infty$ down to $n_f = 3$ :

\Delta E = 13.6\left(\frac{1}{9} - 0\right) = 1.511\ \text{eV}

\lambda = \frac{1240}{1.511} = 820\ \text{nm}

820 nm is longer than 656 nm, so the two series do not overlap: there is a clean gap between them. This is general: because the series limit of one series always sits at longer wavelength than the longest line of the series below it, the hydrogen spectrum sorts itself into separated blocks. That is what makes a barcode identifiable.

Figure 3. Where the three lowest series actually sit on the wavelength axis, drawn on a logarithmic scale. Notice that each series crowds up against its own short-wavelength limit, and that the blocks never touch.

Quick check

A sample of hydrogen is excited to n = 4. How many distinct spectral lines can the sample produce as the atoms cascade down?

Pick one option first.

When the ladder is not quite 13.6

Three corrections turn up often enough that you should recognise them.

Nuclear charge. Already covered: multiply the whole ladder by $Z^2$ and divide the radii by $Z$ .

Reduced mass. The derivation assumed the nucleus stays put. It does not quite; both particles orbit their common centre of mass. The fix is to replace the electron mass $m$ by the reduced mass

\mu = \frac{m M}{m + M} = \frac{m}{1 + m/M}

where $M$ is the nuclear mass. Since the energy is proportional to the orbiting mass, every level scales by $\mu/m$ . For hydrogen $m/M = 1/1836$ , a 0.05% correction: small, but it is exactly what separates hydrogen's spectrum from deuterium's, and that difference is how deuterium was discovered.

A different orbiting particle. If a muon, 207 times heavier than an electron, replaces the electron, the ladder scales by 207: the ground state drops to $207 \times 13.6 = 2815$ eV, about 2.82 keV, and the orbit shrinks by the same factor of 207. Positronium, an electron bound to a positron, has $M = m$ , so $\mu = m/2$ and the ground state is $13.6/2 = 6.8$ eV.

Where this goes wrong

The photon energy is a difference between two levels, never a level itself. A drop from $n = 3$ to $n = 2$ gives 1.89 eV, not 1.51 eV. Compute both energies, subtract, then convert with $\lambda = 1240/\Delta E$ . The same trap runs backwards: an atom that absorbs a 12.09 eV photon from the ground state is not going to the "12.09 eV level", it is going to $-13.6 + 12.09 = -1.51$ eV, which is $n = 3$ .

Worked example 3: a helium line disguised as hydrogen

A He⁺ ion emits a photon of exactly the same wavelength as hydrogen's H-alpha line, and the transition ends on $n_f = 4$ . What is the upper level?

From example 1, the photon energy is 1.889 eV. For He⁺, $Z = 2$ , so the prefactor is $13.6 \times 4 = 54.4$ eV:

1.889 = 54.4\left(\frac{1}{16} - \frac{1}{n_i^2}\right)

Divide: $1.889/54.4 = 0.03472$ . Then

\frac{1}{n_i^2} = 0.0625 - 0.03472 = 0.02778

so $n_i^2 = 1/0.02778 = 36$ and $n_i = 6$ . The coincidence is not an accident. Because the He⁺ ladder is exactly four times hydrogen's, the He⁺ transition $2n_i \to 2n_f$ always lands on hydrogen's $n_i \to n_f$ line. Doubling 3 and 2 gives 6 and 4, which is what you found. Astronomers had to untangle exactly this confusion in early stellar spectra.

Worked example 4: how hard must you hit an atom?

A ground-state hydrogen atom flies at a second ground-state hydrogen atom at rest. What is the minimum kinetic energy the incoming atom needs to excite one of them to $n = 2$ ?

The naive answer is 10.2 eV, the $1 \to 2$ gap. It is wrong, and the reason is momentum. The incoming atom carries momentum, and momentum is conserved, so the pair cannot end up at rest. Whatever kinetic energy they keep is unavailable for exciting anything.

The cheapest outcome is the one where they keep as little as possible, which is when both atoms end up moving together with the same velocity: a perfectly inelastic collision, the same idea you used in collisions. Let each atom have mass $M$ and let the incoming one have speed $v$ . Momentum gives the common final speed:

Mv = 2Mv_f \quad \Rightarrow \quad v_f = \tfrac12 v

Initial kinetic energy is $\tfrac12 Mv^2$ . Final kinetic energy is

\tfrac12 (2M) v_f^2 = M\left(\tfrac{v}{2}\right)^2 = \tfrac14 Mv^2

which is exactly half of what you started with. So only half the incoming kinetic energy is ever available to be turned into internal energy. Setting that half equal to the 10.2 eV you need:

\tfrac12 K_{\min} = 10.2\ \text{eV} \quad \Rightarrow \quad K_{\min} = 20.4\ \text{eV}

You have to hit it twice as hard as the gap suggests. The same bookkeeping runs in reverse when an atom emits: the atom recoils, so the photon gets slightly less than the full gap. The recoil energy is $E_\gamma^2/(2Mc^2)$ , and for a 10.2 eV photon off a hydrogen atom whose rest energy is 938 MeV that is $10.2^2/(2 \times 938 \times 10^6) = 5.5 \times 10^{-8}$ eV. Utterly negligible here, but the same term is the whole story in nuclear gamma emission.

Worth remembering

For very large $n$ the gap between neighbours is tiny and you can get it by differentiating. Since $E_n = -R/n^2$ , a step of one level changes the energy by roughly $dE/dn = 2R/n^3$ . For $n = 100$ that is $2 \times 13.6/10^6 = 2.7 \times 10^{-5}$ eV, giving $\lambda \approx 1240/2.7\times10^{-5} \approx 4.6 \times 10^7$ nm, about 4.5 cm: radio waves from a single atom. These giant Rydberg atoms are real and radio astronomers detect them.

What the model gets wrong

Keep the ladder, distrust the orbits. The energies $E_n = -13.6\,Z^2/n^2$ are exactly what full quantum mechanics gives, and they are right. Everything else in the picture is scaffolding.

The model says nothing about how bright a line is, so it cannot tell you which transitions dominate. It gives no reason the orbits are stable despite the electron accelerating. It fails completely for any atom with more than one electron. And the crisp circular path itself is fiction: the electron has no trajectory, and in the true ground state its angular momentum is zero, not $\hbar$ . Bohr's rule gets the right answer for the wrong reason, which happens more often in physics than anyone likes to admit.

Recognising a Bohr problem

Reach for this machinery when you see any of these:

Integers attached to an atom. " $n = 3$ to $n = 2$ ", "the $n = 4$ level", "the upper state". Quantum numbers in the problem statement mean the ladder.
A wavelength or a photon energy together with an element. Emission lines, absorption lines, series names (Lyman, Balmer, Paschen), "series limit", "ionisation energy". Convert with $\lambda = 1240/\Delta E$ nm and hunt for the pair of integers that fits.
One electron and a nucleus of charge $Ze$ . He⁺, Li²⁺, Be³⁺, or "hydrogen-like". Scale by $Z^2$ .
A substituted particle. Muonic hydrogen, positronium, deuterium versus hydrogen, "the nucleus is not infinitely heavy". Reduced mass.
A collision or a recoil around an excitation. Franck-Hertz tubes, atoms striking atoms, an atom emitting while free. Conserve momentum first, then spend what energy is left on the level gap.

The standard opening move is the same every time: write down $E_{n_i}$ and $E_{n_f}$ as numbers, subtract, and only then decide whether you need a wavelength, a speed, or a radius. Almost every mistake in this topic comes from skipping straight to a formula for $\lambda$ without first writing the two levels.

Bohr's rule was a guess with nothing behind it. Ten years later de Broglie supplied the reason, and it turned out that $L = n\hbar$ is just the statement that a wave has to fit around the circle a whole number of times. That is next.

Problems

Work these on paper, then check your answer here.

Energy of the n = 2 level

EasyWritten for this guideenergy-levels

Hydrogen levels are given by $E_n = -13.6\ \text{eV}/n^2$ . What is the energy of the $n = 2$ level, in eV?

Ionization energy of hydrogen

EasyWritten for this guideionization

How much energy is needed to remove the electron from a ground-state hydrogen atom, in eV?

H-alpha wavelength

EasyWritten for this guidebalmer

A hydrogen atom drops from $n = 3$ to $n = 2$ . What is the wavelength of the emitted photon, in nm? Use $hc = 1240$ eV $\cdot$ nm.

Radius of the third orbit

EasyWritten for this guideorbit-radius

The Bohr radius is $a_0 = 0.0529$ nm and $r_n = n^2 a_0$ . What is the radius of the $n = 3$ orbit in hydrogen, in nm?

Ground state of singly ionized helium

EasyWritten for this guidehydrogen-like

For a hydrogen-like ion, $E_n = -13.6\,Z^2/n^2$ eV. What is the ground-state energy of He $^+$ ( $Z = 2$ ), in eV?

Lyman-alpha photon

EasyWritten for this guidelyman

A hydrogen atom falls from $n = 2$ to $n = 1$ . What is the wavelength of the emitted photon, in nm? Use $hc = 1240$ eV $\cdot$ nm.

Quantized angular momentum

EasyWritten for this guideangular-momentum

Bohr's postulate is $L = n\hbar$ with $\hbar = 1.055\times 10^{-34}$ J $\cdot$ s. What is the orbital angular momentum of an electron in the $n = 4$ orbit, in J $\cdot$ s?

J s

Balmer series limit

EasyWritten for this guidebalmerseries-limit

What is the shortest wavelength in the Balmer series of hydrogen, in nm? Use $hc = 1240$ eV $\cdot$ nm.

Doubly ionized lithium

EasyWritten for this guidehydrogen-like

What is the energy of the $n = 2$ level of Li $^{2+}$ ( $Z = 3$ ), in eV?

10.

Exciting from the ground state

EasyWritten for this guideexcitation

How much energy must a ground-state hydrogen atom absorb to reach the $n = 3$ level, in eV?

11.

Orbital speed in the ground state

NormalWritten for this guideorbit-speed

In the Bohr model the orbital speed is $v_n = \alpha c/n$ with the fine-structure constant $\alpha = 1/137$ . Using $c = 3.00\times 10^8$ m/s, what is the electron's speed in the $n = 1$ orbit of hydrogen, in m/s?

m/s

12.

H-gamma line

NormalWritten for this guidebalmer

What is the wavelength of the $n = 5 \to 2$ transition in hydrogen, in nm? Use $hc = 1240$ eV $\cdot$ nm.

13.

Which level does the atom reach?

NormalWritten for this guideexcitation

A ground-state hydrogen atom absorbs a photon of exactly $12.75$ eV. What is the principal quantum number of the level it reaches?

unitless

14.

Counting spectral lines

NormalWritten for this guidespectracounting

A sample of hydrogen atoms is excited to the $n = 4$ level. How many distinct spectral lines can appear as the atoms cascade down?

unitless

15.

Frequency of the electron's orbit

NormalWritten for this guideorbit

In the $n = 1$ Bohr orbit the electron moves at $2.19\times 10^6$ m/s on a circle of radius $5.29\times 10^{-11}$ m. How many revolutions does it make per second?

1/s

16.

Franck-Hertz photons from mercury

NormalWritten for this guidefranck-hertz

In a Franck-Hertz tube, mercury vapor shows current dips every $4.90$ V. What is the wavelength of the light emitted when the excited mercury atoms decay, in nm? Use $hc = 1240$ eV $\cdot$ nm.

17.

Potential energy in the ground state

NormalWritten for this guidevirial

A ground-state hydrogen atom has total energy $-13.6$ eV. Using the virial theorem for a Coulomb potential, what is the electron's potential energy, in eV?

18.

Muonic hydrogen★

NormalWritten for this guideexotic-atom

A muon ( $207$ times the electron mass) replaces the electron in a hydrogen atom. Treating the proton as infinitely heavy, what is the ground-state binding energy of muonic hydrogen, in keV?

keV

19.

Positronium ground state

NormalWritten for this guidereduced-massexotic-atom

Positronium is a bound electron-positron pair. Using the reduced mass, what is the magnitude of its ground-state binding energy, in eV?

20.

Identifying an absorption line

NormalWritten for this guideabsorption

A stellar spectrum shows hydrogen absorption at $486$ nm out of the $n = 2$ level. What is the upper principal quantum number of the transition? Use $hc = 1240$ eV $\cdot$ nm.

unitless

21.

Paschen series longest line

NormalWritten for this guidepaschen

What is the longest wavelength in the Paschen series (transitions ending on $n = 3$ ) of hydrogen, in nm? Use $hc = 1240$ eV $\cdot$ nm.

22.

Rydberg constant from the model

NormalWritten for this guiderydberg

The Rydberg formula is $1/\lambda = R(1/n_f^2 - 1/n_i^2)$ . Using $13.6$ eV for the Rydberg energy and $hc = 1240$ eV $\cdot$ nm, what is $R$ in m $^{-1}$ ?

1/m

23.

He+ line matching H-alpha★

HardWritten for this guidehydrogen-likescaling

A He $^+$ ion ( $Z = 2$ ) emits a photon of exactly the same wavelength as hydrogen's $n = 3 \to 2$ line. The transition ends on $n_f = 4$ . What is the upper level $n_i$ ?

unitless

24.

Recoil energy on emission

HardWritten for this guiderecoilmomentum

A hydrogen atom at rest emits a $10.2$ eV Lyman-alpha photon. Taking the atomic rest energy as $938$ MeV, how much kinetic energy does the recoiling atom carry, in eV?

25.

Magnetic field at the proton

HardWritten for this guideorbitmagnetism

Treat the $n = 1$ electron as a point charge circling at $2.19\times 10^6$ m/s at radius $5.29\times 10^{-11}$ m. What magnetic field does it produce at the nucleus, in T? Use $\mu_0/4\pi = 1.00\times 10^{-7}$ T $\cdot$ m/A and $e = 1.602\times 10^{-19}$ C.

26.

Threshold for collisional excitation★

HardWritten for this guidecollisionthreshold

A ground-state hydrogen atom moving freely strikes a second ground-state hydrogen atom at rest. What is the minimum kinetic energy of the incoming atom that can excite one atom to $n = 2$ , in eV?

27.

Deuterium isotope shift

HardWritten for this guidereduced-massisotope

Hydrogen's H-alpha line sits at $656.47$ nm. Deuterium's nucleus is heavier, with $m_e/m_d = 1/3670.5$ versus $m_e/m_p = 1/1836.15$ . What is the wavelength shift $\lambda_H - \lambda_D$ , in nm?

28.

Do Balmer and Paschen overlap?

HardWritten for this guideseriesspectra

The Balmer series runs from $365$ nm to $656$ nm. What is the shortest wavelength of the Paschen series, in nm? Use $hc = 1240$ eV $\cdot$ nm.

29.

A Rydberg atom★

Very HardWritten for this guiderydberg-atomcorrespondence

A hydrogen atom is prepared in the $n = 100$ level. What is the wavelength of the photon emitted in the $n = 100 \to 99$ transition, in cm? Use $hc = 1240$ eV $\cdot$ nm.

30.

A gravitationally bound atom

Very HardWritten for this guidescalinggravity

Suppose an electron and a proton were bound only by gravity. Repeating the Bohr derivation with $Gm_em_p$ in place of $ke^2$ gives $r_1 = \hbar^2/(Gm_e^2m_p)$ . Evaluate this radius, in meters. Use $\hbar = 1.055\times 10^{-34}$ J $\cdot$ s, $G = 6.674\times 10^{-11}$ , $m_e = 9.109\times 10^{-31}$ kg, $m_p = 1.673\times 10^{-27}$ kg.

Solutions

1. Energy of the n = 2 level Answer: -3.4 eV

Substitute $n = 2$ : $E_2 = -13.6/4 = -3.40$ eV.

2. Ionization energy of hydrogen Answer: 13.6 eV

Ionization takes the electron from $E_1 = -13.6$ eV to $E = 0$ at rest far away. The required energy is $13.6$ eV.

3. H-alpha wavelength Answer: 656 nm

The energy released is $13.6(1/4 - 1/9) = 1.889$ eV. Then $\lambda = 1240/1.889 = 656$ nm, the red H-alpha line.

4. Radius of the third orbit Answer: 0.476 nm

With $n = 3$ , $r_3 = 9(0.0529) = 0.476$ nm.

5. Ground state of singly ionized helium Answer: -54.4 eV

With $Z = 2$ and $n = 1$ , $E_1 = -13.6(4) = -54.4$ eV.

6. Lyman-alpha photon Answer: 122 nm

The transition energy is $13.6(1 - 1/4) = 10.2$ eV, so $\lambda = 1240/10.2 = 122$ nm. This ultraviolet line is Lyman-alpha.

7. Quantized angular momentum Answer: 4.22 × 10^-34 J s

Directly, $L = 4\hbar = 4(1.055\times 10^{-34}) = 4.22\times 10^{-34}$ J $\cdot$ s.

8. Balmer series limit Answer: 365 nm

The series limit corresponds to $n = \infty \to 2$ , an energy of $13.6/4 = 3.40$ eV. Then $\lambda = 1240/3.40 = 365$ nm.

9. Doubly ionized lithium Answer: -30.6 eV

Using $E_n = -13.6Z^2/n^2$ with $Z = 3$ , $n = 2$ : $E = -13.6(9)/4 = -30.6$ eV.

10. Exciting from the ground state Answer: 12.1 eV

The gap is $E_3 - E_1 = -13.6/9 + 13.6 = 13.6(1 - 1/9) = 12.1$ eV.

11. Orbital speed in the ground state Answer: 2.19 × 10^6 m/s

For $n = 1$ , $v = \alpha c = (3.00\times 10^8)/137 = 2.19\times 10^6$ m/s. That is under 1% of $c$ , which justifies the non-relativistic treatment.

12. H-gamma line Answer: 434 nm

The photon energy is $13.6(1/4 - 1/25) = 2.856$ eV, so $\lambda = 1240/2.856 = 434$ nm, the blue-violet H-gamma line.

13. Which level does the atom reach? Answer: 4 unitless

Solve $13.6(1 - 1/n^2) = 12.75$ , giving $1/n^2 = 0.0625$ and $n^2 = 16$ . The atom lands in $n = 4$ .

14. Counting spectral lines Answer: 6 unitless

Every pair of levels from $1$ to $4$ gives one line, so the count is $\binom{4}{2} = 6$ : the transitions $4\to3, 4\to2, 4\to1, 3\to2, 3\to1, 2\to1$ .

15. Frequency of the electron's orbit Answer: 6.59 × 10^15 1/s

The orbital frequency is $f = v/(2\pi r) = (2.19\times 10^6)/[2\pi(5.29\times 10^{-11})] = 6.59\times 10^{15}$ Hz.

16. Franck-Hertz photons from mercury Answer: 253 nm

Each dip means electrons lost $4.90$ eV exciting a mercury atom, so the excited state lies $4.90$ eV above the ground state. The emitted photon has $\lambda = 1240/4.90 = 253$ nm, in the ultraviolet.

17. Potential energy in the ground state Answer: -27.2 eV

For an inverse-square force the virial theorem gives $\langle U\rangle = 2E$ and $\langle K\rangle = -E$ . Hence $U = 2(-13.6) = -27.2$ eV, with $K = +13.6$ eV.

18. Muonic hydrogen Answer: 2.82 keV

Bohr energies scale linearly with the orbiting particle's mass, $E_1 \propto m$ . So $E_1 = 207 \times 13.6\ \text{eV} = 2.82\times 10^3$ eV $= 2.82$ keV, and the orbit shrinks by the same factor of $207$ .

19. Positronium ground state Answer: 6.8 eV

For equal masses the reduced mass is $\mu = m_e/2$ , and Bohr energies scale with $\mu$ . So the binding energy is $13.6/2 = 6.80$ eV.

20. Identifying an absorption line Answer: 4 unitless

The photon energy is $1240/486 = 2.55$ eV. Setting $13.6(1/4 - 1/n^2) = 2.55$ gives $1/n^2 = 0.0625$ , so $n = 4$ .

21. Paschen series longest line Answer: 1880 nm

The longest wavelength comes from the smallest gap, $n = 4 \to 3$ : $13.6(1/9 - 1/16) = 0.661$ eV. So $\lambda = 1240/0.661 = 1.88\times 10^3$ nm, in the infrared.

22. Rydberg constant from the model Answer: 1.1 × 10^7 1/m

Comparing $hc/\lambda = 13.6(1/n_f^2 - 1/n_i^2)$ with the Rydberg formula gives $R = 13.6/hc$ . Numerically $R = 13.6/1240\ \text{nm}^{-1} = 0.01097\ \text{nm}^{-1} = 1.10\times 10^7$ m $^{-1}$ .

23. He+ line matching H-alpha Answer: 6 unitless

Hydrogen's H-alpha energy is $13.6(1/4 - 1/9) = 1.889$ eV. For He $^+$ , $4(13.6)(1/16 - 1/n_i^2) = 1.889$ gives $1/n_i^2 = 0.02778$ , so $n_i = 6$ . The $Z = 2$ spectrum reproduces the hydrogen lines whenever both quantum numbers are doubled.

24. Recoil energy on emission Answer: 5.54 × 10^-8 eV

Momentum conservation gives the atom $p = E_\gamma/c$ , so $K = p^2/2M = E_\gamma^2/(2Mc^2)$ . That is $(10.2)^2/(2\times 9.383\times 10^8) = 5.54\times 10^{-8}$ eV, utterly negligible compared with the line energy.

25. Magnetic field at the proton Answer: 12.5 T

The Biot-Savart field of a moving point charge at the center of its orbit is $B = (\mu_0/4\pi)\,ev/r^2$ . So $B = 10^{-7}(1.602\times 10^{-19})(2.19\times 10^6)/(5.29\times 10^{-11})^2 = 12.5$ T.

26. Threshold for collisional excitation Answer: 20.4 eV

Momentum must be conserved, so the two equal-mass atoms cannot both stop; at threshold they move together with half the initial speed, retaining half the kinetic energy. Only half the incoming energy is available for excitation, so $K_{\min} = 2 \times 10.2 = 20.4$ eV.

27. Deuterium isotope shift Answer: 0.179 nm

Level energies scale with the reduced mass $\mu = m_e/(1 + m_e/M)$ , so $\lambda \propto 1/\mu$ and $\lambda_D/\lambda_H = \mu_H/\mu_D = (1+1/3670.5)^{-1}(1+1/1836.15)$ . This gives $\lambda_D = 656.29$ nm and a shift of $0.179$ nm, which is how deuterium was discovered.

28. Do Balmer and Paschen overlap? Answer: 820 nm

The Paschen series limit is $n = \infty \to 3$ , with energy $13.6/9 = 1.511$ eV. Then $\lambda = 1240/1.511 = 820$ nm, so the Paschen series lies entirely beyond the Balmer series and the two do not overlap.

29. A Rydberg atom Answer: 4.49 cm

The energy gap is $13.6(1/99^2 - 1/100^2) = 2.761\times 10^{-5}$ eV. Then $\lambda = 1240/2.761\times 10^{-5} = 4.49\times 10^7$ nm $= 4.49$ cm, a microwave photon. Such transitions are how Rydberg atoms are detected.

30. A gravitationally bound atom Answer: 1.2 × 10^29 m

Substituting, the numerator is $\hbar^2 = 1.112\times 10^{-68}$ and the denominator is $Gm_e^2m_p = 9.26\times 10^{-98}$ . The ratio is $1.20\times 10^{29}$ m, roughly $10^{13}$ light years, which shows just how weak gravity is compared with the Coulomb force.