IGCSE, Edexcel, IB and AP Science Courses | Dr Dabour Hub Academy

Overview

🌳 Concept tree

Lesson 1→Hydrogen Atomic Emission Spectrum and Production of X-Rays

├── Why a new model was needed

├── Bohr's postulates (hydrogen atom)

├── Radii and energies of the hydrogen atom

├── Excitation and emission in hydrogen

├── The series of the hydrogen spectrum

├── Limitations

├── Worked examples (Lesson 1)

├── The spectrometer

├── Types of spectra

├── Worked examples (Lesson 2)

├── What are X-rays?

├── Properties of X-rays

├── Production→the Coolidge tube

├── The X-ray spectrum

├── Origin of the characteristic radiation

├── Origin of the continuous radiation (bremsstrahlung)
├── Comparison of characteristic and continuous X-rays

├── Applications of X-rays

├── Worked examples (Lesson 3)

├── Going deeper — how Bohr's numbers are obtained, and a problem-solving workshop
├── Derivation of the radius and energy (hydrogen)

├── Understanding the energy-level diagram

├── Ionisation, excitation and de-excitation

├── More worked examples

├── X-ray quantities for practice

├── Exam-style reasoning guide

├── Chapter summary

└── Common mistakes

Lesson 1: Hydrogen Atomic Emission Spectrum and Production of X-Rays · 1 / 30

Why a new model was needed

Rutherford’s picture of the atom (a tiny positive nucleus surrounded by orbiting electrons) had a serious problem.
According to classical electromagnetism, an electron moving round the nucleus is accelerating, so it should continuously radiate energy, spiral into the nucleus in a tiny fraction of a second, and emit light of continuously changing frequency.
Real atoms are stable, and they emit light only at certain definite frequencies (line spectra).
Bohr resolved the difficulty by adding quantum ideas to the nuclear atom.

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Bohr’s postulates (hydrogen atom)

At the centre of the atom there is a positively charged nucleus.
Negatively charged electrons move around the nucleus in shells
- (loosely called orbits).
- Each shell has a definite energy.
- While an electron stays in one shell it does not radiate energy (these are the "stationary states").
the number of electrons round the nucleus equals the number of positive charges (protons) in the nucleus.
if an electron jumps from an outer shell of higher energy E₂ to an inner shell of lower energy E₁ (E₂ > E₁), the energy difference is emitted as one photon: h ν = E₂ − E₁ Conversely, an electron can jump from a lower to a higher level by absorbing a photon whose energy is exactly equal to the difference of the two levels. A photon with a different energy is not absorbed.
Inside the atom both electric (Coulomb) forces and mechanical (Newton) laws operate: the electrostatic attraction supplies the centripetal force of the electron’s circular motion.
the shell radius is found by assuming that the wave accompanying the electron forms a standing wave round the orbit. This happens only if the circumference of the orbit contains a whole number of wavelengths: n λ = 2π rn, where n = 1, 2, 3, … and λ = h/(m v) is the de Broglie wavelength. Therefore m v rn = n h/2π, and only particular radii are possible.

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Radii and energies of the hydrogen atom

Combining the standing-wave condition with Coulomb’s law and Newton’s second law gives:

Comparison aspect	Main idea	Keywords
The radius of the n-th orbit	rn = n² r₁ with r₁ = 0.053 nm (the radius of the first orbit). So r₂ = 4r₁ r₃ = 9r₁ and so on: the orbits get farther apart as n increases	firstorbitsfartherapart
The energy of the electron in the n-th level	En = − 13.6 / n² eV (n = 1 …) The negative sign means the electron is bound to the nucleus. The energy is lowest (most negative) for n = 1. As n increases the energy rises toward zero. At n = ∞ the electron is free (E = 0)	negativesignmeansbound
•	Energies of the first levels	Energiesfirstlevels

Level	Name	n	E_n (eV)
ground state	K	1	−13.6
first excited	L	2	−3.40
second excited	M	3	−1.51
third excited	N	4	−0.85
fourth excited	O	5	−0.54
free electron		∞	0

Ionisation energy→of hydrogen: the energy needed to remove the electron from the ground state completely = 13.6 eV (≈ 2.18 × 10⁻¹⁸ J). From the n = 2 level it is only 3.4 eV.

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Excitation and emission in hydrogen

When hydrogen gas is excited (for example by an electric discharge), not all atoms are excited in the same way. In different atoms the electron is raised from the first level K (n = 1) to different higher levels (n = 2, 3, 4, …). A single atom’s electron makes one jump at a time; the many atoms together give all possible jumps.
The electron remains in an excited level only for a very short time, the lifetime (about 10⁻⁸ s), and then returns, directly or by steps, to the lowest level (the ground state).
During the fall from level n_i to level n_f the atom emits a photon with
h ν = E_i − E_f = 13.6 (1/n_f² − 1/n_i²) eV, with wavelength λ = hc/(E_i − E_f).
1/λ = RH (1/nf² − 1/ni²), RH = 1.096 × 10⁷ m⁻¹.
Each possible jump has one definite energy difference and therefore one definite frequency and wavelength. That is why the spectrum of hydrogen consists of lines, not a continuous band.
An atom with N levels excited can show N(N−1)/2 different lines (for four levels, six lines).

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The series of the hydrogen spectrum

The lines are grouped into series, according to the level to which the electron falls (the final level n_f):

Series	Final level n_f	Region	Notes
Lyman	1 (K)	ultraviolet	largest energies shortest wavelengths highest frequencies
Balmer	2 (L)	visible (and near UV)	the only series in the visible region
Paschen	3 (M)	infrared
Brackett	4 (N)	infrared
Pfund	5 (O)	far infrared	longest wavelengths, lowest frequencies

Within each series the first line (called α) comes from n_i = n_f + 1 and has the longest wavelength of that series; the lines get closer together and approach the series limit (from n_i = ∞), which is the shortest wavelength of the series.
Typical values:

Comparison aspect	Main idea	Keywords
Lyman first line (2→1)	ΔE = 10.2 eV λ = 121.6 nm. Limit: 13.6 eV λ = 91.2 nm	Limit
Balmer lines	Hα (3→2) ΔE = 1.89 eV λ = 656 nm (red) Hβ (4→2) 2.55 eV 486 nm (blue-green) Hγ (5→2) 2.86 eV 434 nm (violet) Hδ (6→2) 410 nm. Series limit 3.4 eV 365 nm	bluegreenvioletSeries
Paschen first line (4→3)	0.66 eV 1876 nm	1876
Why does Lyman’s series have higher energy than Pfund’s?	The Lyman photons are released when electrons fall to the lowest level n = 1 the deepest level so the energy gaps are the largest (10.2 to 13.6 eV). Pfund photons come from jumps that end on n = 5 where all the levels are close together so the gaps are only about 0.3 to 0.54 eV	photonsreleasedelectronsfall

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Limitations

Bohr’s model works well for hydrogen (and one-electron ions), but cannot be applied directly to multi-electron atoms.
It was later replaced by the wave (quantum) mechanics of Schrödinger, in which the electron is described by a wave function.

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Worked examples (Lesson 1)

Example 1.1 — Wavelength of Balmer’s Hα. Find the energy and wavelength of the photon emitted when an electron falls from n = 3 to n = 2.
E₃ = −13.6/9 = −1.51 eV; E₂ = −13.6/4 = −3.40 eV. ΔE = 1.89 eV.
λ = 1240/1.89 = 656 nm (red, visible: Balmer series).
Check with Rydberg 1/λ = 1.096 × 10⁷ (1/4 − 1/9) = 1.522 × 10⁶ m⁻¹ → λ = 657 nm ✓.
Example 1.2 — Absorption. What photon energy and wavelength are needed to raise the electron of a ground-state hydrogen atom to n = 4?
ΔE = E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV. λ = 1240/12.75 = 97.3 nm (ultraviolet).
A 10 eV photon would not be absorbed (it matches no allowed jump), and a 12.75 eV photon is.

Example 1.3 — Which transition? A line of wavelength 486 nm is observed in the hydrogen spectrum. Find the transition.
E = 1240/486 = 2.55 eV.
The visible region means Balmer’s series (nf = 2): Ei = −3.4 + 2.55 = −0.85 eV → ni = 4.
So the transition is 4 → 2.
Example 1.4 — Longest wavelength of Paschen. n_i = 4, n_f = 3: 1/λ = 1.096 × 10⁷ (1/9 − 1/16) = 5.33 × 10⁵ m⁻¹ → λ = 1.88 μm (infrared).
Example 1.5 — Orbit radius. r₃ = 3² × 0.053 = 0.477 nm. The orbit of n = 3 is nine times as far from the nucleus as that of n = 1.

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The spectrometer

To obtain a pure spectrum (the different colours separated without overlap) we use a spectrometer. It has three main parts:

Comparison aspect	Main idea	Keywords
Collimator	a light source in front of a slit whose width can be adjusted by a screw. The slit is located at the focal point of a convex lens so that parallel rays leave the collimator	lightsourcefrontslit
Turntable carrying a prism.	The prism is set at the position of minimum deviation for the light being studied. At this position the spectrum is sharpest and brightest	positionminimumdeviationlight
Telescope	consisting of two convex lenses (objective and eyepiece). The objective focuses all rays of the same colour at one point of its focal plane giving a sharp image of the slit for each wavelength (each is a "line"). The eyepiece magnifies these images	consistingconvexlensesobjective

Because a spectrometer can also be used for infrared and ultraviolet if suitable materials and detectors are used, it decomposes light into its components, visible and invisible.
Astronomers use spectrometers to find the composition and even the temperature of stars and their gases, because each element has its own characteristic lines.

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Types of spectra

We can classify the spectra in two ways.
By distribution:

Comparison aspect	Main idea	Example	Keywords
Continuous spectrum	contains all wavelengths in a continuous unbroken way (a rainbow band)	from a glowing solid like a filament or the Sun’s photosphere. It depends on temperature not on the material	containswavelengthsunbrokenrainbow
Line spectrum	only certain separated wavelengths (bright lines on a dark background) as from excited atoms of a gas. The lines are characteristic of the element so line spectra serve as "fingerprints" of the elements	—	certainseparatedwavelengthsbright
•	By origin	—	origin

Comparison aspect	Main idea	Keywords
Emission spectrum	produced when excited atoms fall from higher to lower levels and emit photons	producedexcitedatomsfall
Absorption spectrum	when white (continuous) light passes through a cool gas the atoms of the gas absorb photons of exactly those energies that correspond to their allowed jumps. The transmitted light has dark lines at positions where wavelengths are missing. These missing wavelengths are exactly the wavelengths that the same gas emits in its emission spectrum. Fraunhofer lines in the solar spectrum are an absorption spectrum: the cooler gases in the Sun’s atmosphere (mainly hydrogen and helium among other elements) absorb these wavelengths from the Sun’s continuous radiation. Emission line spectrum Absorption spectrum Source hot (excited) gas cool gas in front of a continuous source Appearance bright lines on dark background dark lines on a bright continuous background Cause electrons fall to lower levels electrons rise to higher levels Positions same wavelengths for the same element	whitecontinuouslightpasses

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Worked examples (Lesson 2)

Comparison aspect	Main idea	Keywords
Example 2.1.	A cool hydrogen gas is placed between a white-light source and a spectrometer. Which wavelength is certainly missing from the transmitted light in the visible region if the gas atoms are in the ground state? None: the visible Balmer lines start from n = 2 which is empty in the ground state absorption from n = 1 needs ultraviolet photons (≥ 10.2 eV). Only the UV Lyman lines would be removed. (That is why Fraunhofer’s lines appear only for gases whose atoms are already excited like the hot gases in the Sun’s atmosphere.)	coolhydrogenplacedbetween
Example 2.2.	In a spectrum of a gas the lines are seen at 410 nm 434 nm 486 nm and 656 nm. Which gas and series is this? They match the visible Balmer lines of hydrogen (6→2 5→2 4→2 3→2)	spectrumlinesseenseries

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What are X-rays?

X-rays are invisible electromagnetic waves of very short wavelength (about 10⁻¹¹ to 10⁻⁸ m, i.e. between 0.01 nm and 10 nm), lying in the spectrum between ultraviolet and gamma rays.
They were discovered by the German physicist Wilhelm Röntgen in 1895, who called them "X-rays" (X = unknown) because he did not know what they were.

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Properties of X-rays

They penetrate matter easily, and their penetrating power increases as the wavelength decreases.
They ionize gases.
They are diffracted by crystals (the spacing between the atoms of a crystal is comparable to the wavelength of X-rays).
They affect photographic plates and cause some substances to fluoresce.
They are not deflected by electric and magnetic fields (they carry no charge).
They travel at the speed of light in vacuum.

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Production: the Coolidge tube

The Coolidge tube is a high-vacuum glass tube containing:

a hot filament (cathode) heated by a low-voltage current, which emits electrons by thermionic emission;
a target (anode), made of a metal with a high melting point such as tungsten, set at a slope in front of the cathode;
a high DC voltage (tens of kilovolts) between filament and target, which accelerates the electrons so that they hit the target with large kinetic energy (eV);
a copper rod with cooling fins, which removes the heat produced (most of the electrons’ energy, over 99 %, is turned to heat; a tiny part becomes X-rays).
When an electron collides with the tungsten target, part or all of its kinetic energy is converted to X-ray photons.

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The X-ray spectrum

Analysing the X-rays by wavelength shows two parts:

Comparison aspect	Main idea	Keywords
A continuous spectrum	containing all wavelengths above a certain minimum value whatever the target material	containingwavelengthsabovecertain
A line spectrum	at definite wavelengths that depend on the target material called the characteristic X-rays these appear as sharp peaks on top of the continuous background	definitewavelengthsdependtarget

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Origin of the characteristic radiation

An incident electron may collide with an electron in an inner shell (close to the nucleus) of a target atom.
If it gives that electron enough energy, the electron jumps to a higher level or leaves the atom, creating a vacancy.
An electron from a higher level falls into the vacancy and the difference in energy is emitted as an X-ray photon of definite wavelength:
hν = hc/λ = ΔE (6-1)
Notes:

The wavelength of the characteristic radiation depends only on the target element, not on the voltage. The larger the atomic number of the target, the larger the energy differences between its inner levels, so the shorter the wavelength.
At low voltages the incoming electrons do not have enough energy to remove inner electrons, so no characteristic lines appear. They appear only when the voltage exceeds a certain value.
Increasing the voltage increases the intensity and shifts the continuous spectrum toward shorter wavelengths, but the position of the characteristic lines does not change.

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Origin of the continuous radiation (bremsstrahlung)

When a fast electron passes near the atoms of the target, the electric fields of the atoms (repulsion by the electrons and attraction by the nucleus) make it slow down.
According to Maxwell and Hertz’s electromagnetic theory a decelerating charge radiates.
This is called braking radiation (German: bremsstrahlung).
The electron may lose its energy in many small steps (many collisions), or in a single step, so the emitted wavelengths cover all values, which is why the spectrum is continuous.
The shortest wavelength occurs when all the electron’s kinetic energy goes into one photon: eV = hc/λmin, so λmin = 1240/V(volts) nm.
This minimum wavelength decreases when the accelerating voltage is increased, and depends on the voltage only.

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Comparison of characteristic and continuous X-rays

Comparison aspect	Continuous (braking) X-rays	Characteristic X-rays
Spectrum	continuous range of wavelengths	sharp lines at certain wavelengths
Origin	deceleration of electrons by target atoms	transitions of inner electrons to fill vacancies
Depends on	accelerating voltage (λ_min)	target element only
Appears	at all voltages	only above a threshold voltage

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Applications of X-rays

Comparison aspect	Main idea	Keywords
Studying crystal structure	the atoms of a crystal form a three-dimensional grating so X-rays are diffracted and give bright and dark fringes (interference) depending on path differences. From the pattern the distances between atoms and the crystal structure are determined. (Bragg father and son developed this.)	atomsformthreedimensional
Industrial radiography	because of their great penetrating power X-rays reveal cracks and defects in metal parts welds and pipes	becausegreatpenetratingpower
Medicine	imaging of bones and fractures the chest the teeth and diagnosing diseases also in treatment of some tumours	imagingbonesfractureschest
Security checks	baggage scanners	baggagescanners
Research	analysing materials (X-ray fluorescence)	analysingmaterialsfluorescence

Safety→X-rays ionize tissue, so exposure must be limited and protection such as lead shields should be used.

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Worked examples (Lesson 3)

Comparison aspect	Main idea	Keywords
Example 3.1 — Cut-off wavelength.	A Coolidge tube operates at 30 kV. Find the shortest wavelength and the highest frequency of the X-rays	CoolidgetubeoperatesFind
•	eV = hc/λ_min → λ_min = 1240/30 000 = 0.0413 nm. ν_max = c/λ_min = 3 × 10⁸/4.13 × 10⁻¹¹ = 7.3 × 10¹⁸ Hz. (Check: ν = eV/h = 1.6 × 10⁻¹⁹ × 3 × 10⁴/6.63 × 10⁻³⁴ = 7.2 × 10¹⁸ Hz.)	λmin12400413νmax
Example 3.2 — Speed of the electrons.	At 25 kV kinetic energy at the target = eV = 4 × 10⁻¹⁵ J speed v = √(2KE/m) = 9.4 × 10⁷ m/s (about 31 % of c so the result is approximate)	kineticenergytargetresult
Example 3.3 — Characteristic line.	An inner-shell transition in a target atom releases 8.0 keV. λ = 1240/8000 = 0.155 nm. If the voltage is raised from 40 kV to 60 kV this wavelength is unchanged	innershelltransitiontarget

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Going deeper — how Bohr’s numbers are obtained, and a problem-solving workshop

This lesson is not a new topic; it shows where the formulas of Lesson 1 come from and trains you to use them in exam-type problems.

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Derivation of the radius and energy (hydrogen)

Take an electron (mass m, charge e) moving in a circle of radius r with speed v round a nucleus of charge +e.
Step 1 — force balance. The Coulomb attraction provides the centripetal force:
K e²/r² = m v²/r, where K = 9 × 10⁹ N·m²/C². Hence m v² = K e²/r. (i)
Step 2 — quantum (standing-wave) condition. Only orbits whose circumference holds a whole number of de Broglie wavelengths are allowed: 2π r = n λ = n h/(m v), so m v r = n h/2π. (ii)
Step 3 — radius. From (ii), v = n h/(2π m r). Substitute into (i): m n² h²/(4π² m² r²) = K e²/r, which gives
r_n = n² h²/(4π² m K e²) = n² r₁, with r₁ = 5.3 × 10⁻¹¹ m = 0.053 nm.
Step 4 — energy. Kinetic energy = ½mv² = K e²/(2r). Potential energy = −K e²/r. The total is
E_n = −K e²/(2 r_n) = −(K e²/2 r₁)/n² = −13.6/n² eV.
(You do not need to reproduce the derivation in the exam, but you should understand what it shows: the energy is quantized because the radius is quantized, and the radius is quantized because the electron wave must fit the orbit.)

Step 5 — speed.From (ii) and r_n = n² r₁, v_n = v₁/n, where v₁ = 2.2 × 10⁶ m/s (about 1/137 of c). The electron moves more slowly in outer orbits.

Check of the energiesfor n = 1, KE = +13.6 eV and PE = −27.2 eV, so E = −13.6 eV. Note that the kinetic energy of a bound electron equals minus the total energy.

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Understanding the energy-level diagram

Draw horizontal lines at −13.6
−3.40
−1.51
−0.85
−0.54 eV and 0 (free). Then:

Downward arrows are emissions. The length of an arrow is proportional to the photon energy (long arrow = ultraviolet, short arrow = infrared).
Upward arrows are absorptions. The photon must match exactly.
All the arrows that end on the same level belong to the same series.
Two different paths between the same pair of levels give the same total energy: a 4 → 1 jump emits 12.75 eV; a 4 → 2 followed by 2 → 1 emits 2.55 + 10.2 = 12.75 eV in two photons.
The ratio of wavelengths for two jumps is the inverse ratio of their energies.

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Ionisation, excitation and de-excitation

Comparison aspect	Main idea	Keywords
Excitation energy	from the ground state to level n: E_n − E₁. For n = 2: 10.2 eV (first excitation energy). For n = 3: 12.09 eV.	groundstatelevelfirst
Ionisation energy	from the ground state = 13.6 eV from level n: 13.6/n² eV	groundstatelevel
•	A free electron colliding with an atom can excite it if its kinetic energy is at least the excitation energy a photon needs exactly the energy difference but a photon with energy larger than 13.6 eV can ionise an atom with the surplus becoming the kinetic energy of the freed electron (like the photoelectric effect with the ionisation energy playing the role of the work function)	freeelectroncollidingatom

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More worked examples

Comparison aspect	Main idea	Keywords
Example 4.1 — Photoionisation.	A photon of 20 eV strikes a hydrogen atom in the ground state. What is the kinetic energy of the freed electron? KE = 20 − 13.6 = 6.4 eV	photonstrikeshydrogenatom
Example 4.2 — Minimum frequency to ionise.	ν = 13.6 × 1.6 × 10⁻¹⁹/6.63 × 10⁻³⁴ = 3.28 × 10¹⁵ Hz λ = 91.2 nm (the Lyman limit)	10¹⁵Lymanlimit
Example 4.3 — Number of lines.	Hydrogen atoms are excited to n = 4. How many different lines can be emitted and is the longest wavelength? Transitions: 4→3 4→2 4→1 3→2 3→1 2→1: six lines. Smallest energy: 4→3 (0.66 eV 1876 nm infrared Paschen)	Hydrogenatomsexcitedmany
Example 4.4 — Wavelength ratio.	Compare the wavelengths of Hα (3→2) and the Lyman α (2→1). Energies are 1.89 eV and 10.2 eV so λ(Hα)/λ(Lyα) = 10.2/1.89 = 5.4. (656 nm / 121.6 nm = 5.4 ✓)	ComparewavelengthsLymanEnergies

Comparison aspect	Main idea	Keywords
Example 4.5 — Series limit of Balmer.	From n = ∞: 1/λ = R_H/4 = 2.74 × 10⁶ m⁻¹ → λ = 365 nm (just in the ultraviolet). All the Balmer lines lie between 365 nm and 656 nm.	justultravioletlinesbetween
Example 4.6 — Time to cross.	An excited electron stays about 10⁻⁸ s. During this time it makes about v₂/(2π r₂) revolutions: with v₂ = 1.1 × 10⁶ m/s and r₂ = 0.212 nm the frequency is 8.2 × 10¹⁴ revolutions per second so about 8 million revolutions before emission. The classical theory would have the electron radiate on every revolution Bohr’s stationary-state idea says it does not	excitedelectronstaysDuring

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X-ray quantities for practice

Comparison aspect	Main idea	Keywords
Cut-off wavelength	λ_min (nm) = 1.24/V(kV). At 62 kV, 0.02 nm; at 124 kV, 0.01 nm.	λmin
Energy of an electron arriving at the target	KE = eV at 50 kV it is 50 keV = 8 × 10⁻¹⁵ J
Percentage of energy turned into X-rays	is small (about 1 %) the rest heats the target hence the cooling fins and the tungsten (high melting point)	smallrestheatstarget
Why tungsten?	High melting point high atomic number (more efficient X-ray production) good thermal conductivity when mounted on copper	Highmeltingpointatomic
Diffraction condition (extension, not required by the book)	for a crystal with plane spacing d constructive interference occurs when 2 d sin θ = m λ. With λ = 0.154 nm and d = 0.282 nm the first-order angle is sin θ = 0.154/0.564 = 0.273 θ = 15.8°	crystalplanespacingconstructive
•	Why can X-rays be diffracted by crystals but not by an ordinary grating? Diffraction is strong only when the wavelength is comparable to the separation of the scatterers. The spacing between atoms in crystals is of the order of 0.1–0.5 nm similar to X-ray wavelengths whereas ordinary gratings have spacing of micrometres	raysdiffractedcrystalsordinary

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Exam-style reasoning guide

Comparison aspect	Main idea	Keywords
Which series?	Look at the final level or at the region: UV → Lyman visible → Balmer IR → Paschen/Brackett/Pfund	Lookfinallevelregion
Which line of the series?	The longest wavelength (smallest energy) is the first line (n_i = n_f + 1); the shortest is the series limit (n_i = ∞).	longestwavelengthsmallestenergy
Energies versus wavelengths	large energy ⇔ high frequency ⇔ short wavelength	largeenergyhighfrequency
Emission versus absorption	emission lowers n absorption raises n	lowersraises
Convert units carefully	E(eV) = 1240/λ(nm)	1240

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Chapter summary

Comparison aspect	Main idea	Keywords
Bohr’s postulates	nucleus electrons in shells with definite energies without radiation hν = Ehigh − Elow quantum condition nλ = 2πr	nucleuselectronsshellsdefinite
Hydrogen	En = −13.6/n² eV rn = n² × 0.053 nm
Hydrogen series	Lyman (UV) Balmer (visible) Paschen Brackett (IR) Pfund (far IR)	LymanBalmervisiblePaschen
Spectrometer	collimator + prism at minimum deviation + telescope. Spectra: continuous / line emission / absorption Fraunhofer lines	collimatorprismminimumdeviation
X-rays	discovered by Röntgen in 1895 produced in a Coolidge tube spectrum = continuous (braking) + characteristic lines of the target used for crystal studies industry and medicine	discoveredRöntgen1895produced

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Common mistakes

A Balmer line is a jump to n = 2, not from n = 2.
Emission lines and absorption lines of the same gas occur at the same wavelengths.
Characteristic X-ray wavelengths depend on the target material, not on the voltage; λ_min depends on voltage.
In energy-level calculations remember the negative sign ΔE = Ehigh − Elow is positive for emission.
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Part 30

📖

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Physics (Thanaweya Amma 2026/2027)

Lesson 1 – Hydrogen Atomic Emission Spectrum and Production of X-Rays
The hydrogen spectrum provided experimental verification for Niels Bohr's atomic model.
When an excited electron transitions to a lower energy level, a photon of energy ΔE = Eupper – Elower = h·ν is emitted across distinct spectral series (Lyman, Balmer, Paschen, Brackett, Pfund).
X-rays are produced in Coolidge tubes via continuous Bremsstrahlung and characteristic line emissions.

🌳 Conceptual Hierarchy Tree

Atomic Spectra and X-Ray Physics

├── 1. Hydrogen Spectral Series

Comparison aspect	Main idea	Keywords
│ ├── Lyman Series (Transitions to n=1)	Ultraviolet region (Highest energy)	UltravioletregionHighestenergy
│ ├── Balmer Series (Transitions to n=2)	Visible light region (Only visible series)	Visiblelightregion
│ ├── Paschen Series (Transitions to n=3)	Near Infrared region	NearInfraredregion
│ ├── Brackett Series (Transitions to n=4)	Mid Infrared region	Infraredregion
│ └── Pfund Series (Transitions to n=5)	Far Infrared region (Lowest energy)	InfraredregionLowestenergy

└── 2. X-Ray Production (Coolidge Tube)

├── Continuous Spectrum (Bremsstrahlung / Braking Radiation):
│ ├── Electrons decelerate near heavy target nuclei; λ_min = (h · c) / (e · V)
│ └── Depends ONLY on the accelerating potential difference V
└── Characteristic Line Spectrum:
├── Incident electron knocks out inner-shell (K or L) electron; outer electron falls into vacancy
└── Depends ONLY on the Atomic Number (Z) of target material (Tungsten)

📋 Master Teacher's Blackboard Secrets

Comparison aspect	Main idea	Keywords
RULE 1	The minimum wavelength of continuous X-rays λ_min = hc / (eV) depends ONLY on the accelerating potential difference V and is completely independent of the target material	minimumwavelengthcontinuousrays
RULE 2	The characteristic X-ray wavelength depends ONLY on the atomic number Z of the target (increasing Z shifts characteristic lines to shorter wavelengths)	characteristicwavelengthdependsatomic
RULE 3	Balmer series is the ONLY hydrogen spectral series visible to the human eye	Balmerserieshydrogenspectral
📊 Matrix	Hydrogen Spectral Series Classification Matrix	HydrogenSpectralSeriesClassification

Series Name	Final Lower Level (n)	Spectral Region	Relative Energy / Frequency
Lyman Series	n = 1	Ultraviolet (UV)	Highest energy / Shortest wavelength
Balmer Series	n = 2	Visible Light (4 lines)	Visible spectrum
Paschen Series	n = 3	Near Infrared (IR)	Low energy
Brackett Series	n = 4	Mid Infrared (IR)	Very low energy
Pfund Series	n = 5	Far Infrared (IR)	Lowest energy / Longest wavelength

Lesson 1: Hydrogen Atomic Emission Spectrum and Production of X-Rays · 30 / 30