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Modern Physics

X-ray Spectra

X-ray spectra consist of a continuous bremsstrahlung background produced by the deceleration of high-speed electrons and characteristic line spectra arising from electronic transitions between inner atomic shells. These high-energy electromagnetic radiations follow Moseley's law, connecting frequency to atomic number.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Continuous and characteristic X-rays, Coolidge tube working, Duane-Hunt law, minimum wavelength, and qualitative aspects of X-ray production.

PGT extension

Moseley's law derivation and application, fine structure of characteristic X-rays, screening constants, absorption edges, and Auger effect.

Prerequisites

Bohr's theory of hydrogen atom and energy levels
Pauli's exclusion principle and inner shell electron configurations
Bremsstrahlung radiation mechanics

Origin of Continuous X-ray Spectra

Produced when high-energy electrons strike a heavy metal target and undergo sudden deceleration (bremsstrahlung) in the Coulomb field of target nuclei.
The energy of the emitted photon ranges from zero up to the maximum kinetic energy of the incident electron, E = eV.
The short-wavelength limit (cut-off wavelength) is independent of the target material and depends solely on the accelerating potential.

Origin of Characteristic X-ray Spectra

Arises when an incident electron knocks out an inner-shell electron (K or L shell), creating a vacancy.
An electron from an outer shell jumps down to fill the vacancy, emitting a photon with energy equal to the difference between the two orbital energy levels.
The spectrum consists of sharp lines grouped into series named K, L, M, N based on the principal quantum number of the shell where the vacancy originated.

Moseley's Law and Atomic Number

H.G.J. Moseley established that the square root of the frequency of a given characteristic X-ray line is directly proportional to the atomic number Z of the target element.
Expressed mathematically as sqrt(nu) = a(Z - b), where a is a proportionality constant and b is the screening constant.
Proved that atomic number, rather than atomic weight, is the fundamental property governing the chemical and physical identity of elements.

Fine Structure and X-ray Notation

Characteristic lines are designated using Siegbahn notation (Ka1, Ka2, Kb1, etc.) or IUPAC notation.
Transitions into the K shell from L_II and L_III subshells give rise to the doublet Ka2 and Ka1, explained by spin-orbit coupling.
Selection rules for X-ray transitions follow quantum mechanics: delta(l) = +-1 and delta(j) = 0, +-1.

Derivations

Duane-Hunt Law for Minimum Wavelength

Conditions: Electron accelerated through potential V striking a target

Equate the maximum kinetic energy of the incident electron to the maximum energy of the emitted X-ray photon.
$K_max = e V = h \nu_{max}$
Substitute frequency relation \n$u_{max} = \frac{c}{\lambda_{min}} into the energy equation$.
e $V = \frac{h c}{\lambda_{min}}$

Result: \lambd$a_{min} = \frac{h c}{e V}$

Moseley's Law Frequency Derivation

Conditions: Transition of an electron between principal quantum numbers n_2 and n_1 in a multielectron atom

Use Bohr's energy expression modified for an effective nuclear charge $Z_{eff} = Z - b$.
$E_n = - \frac{m e^4 (Z - b)^2}{8 \epsilon_0^2 h^2 n^2}$
Write the energy difference for the transition from n_2 to n_1: \Delta $E = h \nu = E_{n2} - E_{n1}$.
$\nu = \frac{m e^4}{8 \epsilon_0^2 h^3} (Z - b)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Take the square root of both sides to obtain a linear relation with (Z - b).

Result: \sqrt{\nu} = \sqrt{ \frac{m e^4}{8 \epsilon_0^2 h^3} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) } (Z - b)

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Minimum X-ray Wavelength (Duane-Hunt Law)	$$\lambda_{min} = \frac{h c}{e V}$$	h = Planck's constant, c = speed of light, e = elementary charge, V = accelerating voltage m	Valid for continuous X-ray spectrum cut-off
Moseley's Law	$$\sqrt{\nu} = a(Z - b)$$	$\nu = frequency of characteristic X-ray$, a = constant depending on transition, Z = atomic number, b = screening constant Hz^{1/2}	Valid for K and L characteristic lines
Energy of Characteristic X-ray Photon	$$h \nu = E_{initial} - E_{final}$$	h = Planck's constant, \nu = frequency, E = orbital energy levels J	Energy conservation during electronic transition
Wavelength in terms of Rydberg constant	$$\frac{1}{\lambda} = R (Z - b)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$	R = Rydberg constant, Z = atomic number, b = screening constant, n_1, $n_2 = principal quantum numbers$ m^{-1}	Inner shell single-electron transition approximation
Power Input to X-ray Tube	$$P = V \times I$$	V = accelerating potential difference, I = tube current W	Total electrical power supplied
Efficiency of X-ray Production	$$\eta = k Z V$$	$k = constant \approx 9 \times 10^{-10} V^{-1}$, Z = atomic number, V = accelerating voltage Dimensionless (fraction or percentage)	Low to medium accelerating voltages
Absorption Coefficient Law	$$I = I_0 e^{-\mu x}$$	$I_0 = initial intensity$, I = transmitted intensity, \mu = linear absorption coefficient, x = thickness m^{-1} for \mu, m for x	Monochromatic X-ray beam passing through homogeneous absorber
Bragg's Law for X-ray Diffraction	$$2 d \sin(\theta) = n \lambda$$	d = interplanar spacing, \theta = Bragg angle, n = order of diffraction, \lambda = wavelength m	Crystalline targets for crystal structure analysis

Graphs and experiments

Continuous and Characteristic X-ray Spectrum

Draw: Intensity (y-axis) vs Wavelength (x-axis) showing a sharp cut-off at lambda_min, a continuous background curve peaking at higher wavelengths, and sharp characteristic peaks superimposed.

Interpret: The sharp cut-off represents maximum photon energy from electron deceleration. Sharp peaks represent discrete electronic transitions in inner shells.

Exam use: Questions test the dependence of lambda_min on accelerating voltage and the independence of characteristic peak positions from voltage.

Moseley's Plot (sqrt(nu) vs Z)

Draw: Straight line plot of sqrt(nu) on the y-axis against atomic number Z on the x-axis for K_alpha lines, with a small positive intercept on the Z axis.

Interpret: Confirms the linear relationship predicted by Moseley's law, validating the atomic number concept over atomic mass.

Exam use: Frequently tested for identifying slopes, screening constants, or finding unknown atomic numbers from frequency data.

Coolidge Tube Experimental Setup

Draw: Evacuated glass bulb with a heated cathode filament, focusing cup, high-voltage DC supply across terminals, and a cooled heavy metal anode (tungsten) inclined at 45 degrees.

Interpret: Demonstrates the conversion of electrical energy into thermal energy and X-ray radiation, highlighting water cooling to prevent target melting.

Exam use: Conceptual questions on the role of filament current (controls intensity) and accelerating voltage (controls hardness/wavelength).

Solved numericals

TGT core

Calculate the minimum wavelength of X-rays produced by an X-ray tube operating at an accelerating potential of 40 kV.

Identify given values: $V = 40 kV = 40 \times 10^3 V$, $h = 6.63 \times 10^{-34} J s$, $c = 3 \times 10^8 m/s$, $e = 1.6 \times 10^{-19} C$.
Use Duane-Hunt law: \lambd$a_{min} = \frac{h c}{e V}$.
Substitute values: \lambd$a_{min} = \frac{(6.63 \times 10^{-34}) (3 \times 10^8)}{(1.6 \times 10^{-19}) (40 \times 10^3)}$.
Calculate result: \lambd$a_{min} = \frac{1.989 \times 10^{-25}}{6.4 \times 10^{-15}} = 3.11 \times 10^{-11} m = 0.311 nm$.

Answer: 0.311 nm (or 0.311 Å)

PGT extension

Using Moseley's law sqrt(\nu) = a(Z - b), find the frequency of the K_alpha line for an element with atomic number Z = 41, given constant $a = 4.95 \times 10^7 Hz^{1/2}$ and screening constant b = 1 for the K shell.

Identify given parameters: Z = 41, b = 1, $a = 4.95 \times 10^7 Hz^{1/2}$.
Calculate effective atomic number: Z - $b = 41 - 1 = 40$.
Calculate square root of frequency: \sqrt{\nu} = (4.95 \times 10^7) \times $40 = 1.98 \times 10^9 Hz^{1/2}$.
Square both sides to find frequency: $\nu = (1.98 \times 10^9)^2 = 3.92 \times 10^{18} Hz$.

Answer: 3.92 \times 10^{18} Hz

PGT extension

In a Coolidge tube, if the accelerating potential is increased from 20 kV to 40 kV, what is the ratio of the minimum wavelength of continuous X-rays produced?

Recall Duane-Hunt law: \lambda_{min} is inversely proportional to accelerating voltage V, i.e., \lambd$a_{min} = \frac{hc}{eV}$.
Set up the ratio: \frac{\lambda_{min1}}{\lambda_{min2}} = \frac{V_2}{V_1}.
Substitute given voltages: $V_1 = 20 kV$ and $V_2 = 40 kV$.
Ratio becomes: \frac{\lambda_{min1}}{\lambda_{min2}} = \frac{40 kV}{20 kV} = \frac{2}{1}.

Answer: 2 : 1

Assumptions and limitations

Moseley's simple screening formula assumes single-electron transitions from the L to K shell, ignoring complex multi-electron screening interactions in higher shells.
The Duane-Hunt limit assumes all kinetic energy of a single incident electron is converted into a single photon in a single collision.

Common misconceptions

Common error: The characteristic X-ray wavelength depends on the accelerating voltage of the tube.

Correct: Characteristic X-ray wavelengths depend only on the target material's atomic structure.

Accelerating voltage only determines whether the inner shell electron can be knocked out; once emitted, the line wavelength is governed by atomic energy levels.

Common error: Continuous X-ray spectra are produced by electronic transitions between discrete atomic energy levels.

Correct: Continuous X-rays are produced by bremsstrahlung radiation due to deceleration of electrons.

Energy loss is continuous during deceleration in the nuclear Coulomb field, unlike discrete transitions.

Quick revision

Minimum wavelength lambd$a_min = hc / (eV) depends solely on the accelerating potential V$.
Minimum X-ray Wavelength (Duane-Hunt Law): \lambd$a_{min} = \frac{h c}{e V}$
Moseley's Law: \sqrt{\nu} = a(Z - b)

Must remember

Continuous X-rays are produced by bremsstrahlung radiation due to sudden deceleration of high-energy electrons in the Coulomb field of target nuclei.
The short-wavelength cut-off (Duane-Hunt law) is given by $\lambda_{min} = \frac{hc}{eV}$, which depends only on accelerating potential and is independent of target material.
Characteristic X-rays arise from electronic transitions filling inner-shell vacancies (K, L, M series) and have wavelengths dependent on target atomic number.
Moseley's Law states that $\sqrt{\nu} = a(Z - b)$, proving that atomic number $Z$ is the fundamental property governing X-ray spectra.
Fine structure of characteristic X-rays (like $K_{\alpha 1}$ and $K_{\alpha 2}$ doublets) is explained by spin-orbit coupling and quantum mechanical selection rules ($\Delta l = \pm 1$, $\Delta j = 0, \pm 1$).

Glossary

BremsstrahlungBraking radiation emitted when high-energy charged particles are decelerated by the electric field of atomic nuclei.

Duane-Hunt LawRelation giving the minimum wavelength of continuous X-rays corresponding to the maximum photon energy equal to the incident electron kinetic energy.

Moseley's LawEmpirical law relating the square root of characteristic X-ray frequency to the atomic number of the target element.

Screening ConstantConstant $b$ in Moseley's law accounting for the shielding of nuclear charge by inner electrons.

Siegbahn NotationTraditional naming convention for characteristic X-ray spectral lines using Greek letters like $K_{\alpha}$, $K_{\beta}$.

Continuous vs. Characteristic X-ray Spectra

Feature	Continuous X-Rays (Bremsstrahlung)	Characteristic X-Rays
Origin	Deceleration of electrons in target nuclear field	Inner-shell electronic transitions
Spectrum Type	Continuous band with a sharp short-wavelength cut-off	Line spectrum consisting of sharp peaks
Dependency on Target ($Z$)	Independent of target material ($Z$)	Strongly dependent on atomic number ($Z$) via Moseley's law
Dependency on Voltage	Cut-off wavelength depends on accelerating voltage ($V$)	Wavelength depends on atomic energy levels, requiring threshold voltage

Practice Quiz

Choose an option to reveal the explanation.

1. What is the physical origin of continuous X-rays (Bremsstrahlung)?

Explanation: Continuous X-rays are produced when high-energy electrons undergo sudden deceleration (bremsstrahlung) in the electric field of target nuclei.

2. The short-wavelength cut-off $\lambda_{min}$ of continuous X-rays depends on:

Explanation: According to the Duane-Hunt law, $\lambda_{min} = \frac{hc}{eV}$, which depends solely on the accelerating potential $V$ and fundamental constants.

3. Characteristic X-ray spectra consist of:

Explanation: Characteristic X-ray spectra consist of sharp lines grouped into series ($K, L, M$) corresponding to discrete energy level transitions.

4. Moseley's law relates the frequency $\nu$ of characteristic X-rays to which property?

Explanation: Moseley's law is given by $\sqrt{\nu} = a(Z - b)$, establishing a direct linear relation with atomic number $Z$.

5. What is the condition for the minimum wavelength of continuous X-rays for an accelerating voltage of $V$?

Explanation: The maximum energy of an emitted photon equals the maximum kinetic energy of the incident electron, $h\nu_{max} = \frac{hc}{\lambda_{min}} = eV$.

6. Which transition gives rise to the $K_{\alpha}$ characteristic X-ray line?

Explanation: The $K_{\alpha}$ line is emitted when an electron drops from the $L$ shell to fill a vacancy in the $K$ shell.

7. The fine structure splitting of characteristic X-ray lines (such as $K_{\alpha 1}$ and $K_{\alpha 2}$) is primarily due to:

Explanation: Transitions from $L_{II}$ and $L_{III}$ subshells into the $K$ shell produce doublets due to spin-orbit coupling.

8. According to quantum mechanical selection rules, X-ray electronic transitions must satisfy:

Explanation: The selection rules for X-ray dipole transitions are $\Delta l = \pm 1$ and $\Delta j = 0, \pm 1$.

9. If the accelerating potential in an X-ray tube is doubled, the minimum wavelength $\lambda_{min}$ will:

Explanation: Since $\lambda_{min} = \frac{hc}{eV}$, wavelength is inversely proportional to potential $V$. Doubling $V$ halves $\lambda_{min}$.

10. Which of the following parameters does NOT affect the wavelength of characteristic X-rays?

Explanation: Accelerating voltage only determines if inner-shell electrons can be knocked out; once emitted, characteristic wavelength depends strictly on atomic energy levels.

11. The efficiency of X-ray production in a Coolidge tube is approximately proportional to:

Explanation: The efficiency formula is given by $\eta = k Z V$, showing direct proportionality with target atomic number $Z$ and voltage $V$.

12. The linear absorption coefficient $\mu$ describes how an X-ray beam intensity decreases according to:

Explanation: The absorption coefficient law is expressed exponentially as $I = I_0 e^{-\mu x}$ for a monochromatic beam through a homogeneous absorber.

13. What happens to the continuous X-ray spectrum when the tube current is increased while keeping the voltage constant?

Explanation: Increasing the tube current increases the number of electrons striking the target per second, increasing the intensity without changing $\lambda_{min}$.

14. Bragg's law for X-ray diffraction in crystals is given by:

Explanation: Bragg's law is $2d \sin\theta = n\lambda$, relating interplanar spacing $d$, Bragg angle $\theta$, and wavelength $\lambda$.

15. Moseley's discovery proved that the fundamental property governing an element's chemical and physical identity is its:

Explanation: Moseley's law proved that atomic number $Z$, rather than atomic weight, is the fundamental property of elements.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. If the accelerating voltage of an X-ray tube is $40\text{ kV}$, the minimum wavelength of the emitted continuous X-rays is approximately:

Explanation: Using $\lambda_{min} = \frac{hc}{eV} = \frac{12400\text{ eV}\cdot\text{\AA}}{40000\text{ V}} = 0.31\text{ \AA} = 0.031\text{ nm}$.

2. The frequency of the $K_{\alpha}$ line for an element of atomic number $Z = 41$ is such that $\sqrt{\nu}$ is proportional to $(41 - b)$. If the screening constant for the $K$ shell is $b = 1$, find the relative ratio of frequencies for $Z = 41$ and $Z = 21$ (with same $b$).

Explanation: By Moseley's law, $\sqrt{\nu} \propto (Z - b)$. Thus $\frac{\sqrt{\nu_1}}{\sqrt{\nu_2}} = \frac{41 - 1}{21 - 1} = \frac{40}{20} = 2$. Squaring both sides gives $\frac{\nu_1}{\nu_2} = 4$.

3. In an X-ray tube, electrons are accelerated through $50\text{ kV}$ and strike a target of $Z = 80$. What is the approximate power input if the tube current is $10\text{ mA}$?

Explanation: Power input $P = V \times I = 50 \times 10^3\text{ V} \times 10 \times 10^{-3}\text{ A} = 500\text{ W}$.

4. The wavelength of the $K_{\alpha}$ X-ray line of an element is $0.71\text{ \AA}$. Assuming Rydberg constant $R = 1.1 \times 10^7\text{ m}^{-1}$ and screening constant $b = 1$ for the $K$ shell ($n_1 = 1, n_2 = 2$), the atomic number $Z$ is approximately:

Explanation: Using $\frac{1}{\lambda} = R(Z-1)^2(\frac{1}{1^2} - \frac{1}{2^2}) = \frac{3}{4} R (Z-1)^2$. Substituting $\lambda = 0.71 \times 10^{-10}\text{ m}$ yields $Z - 1 \approx 41 \implies Z = 42$.

5. If a monochromatic X-ray beam of intensity $I_0$ passes through a sheet of metal of thickness $2\text{ mm}$ and its intensity drops to $I_0 / e$, what is the linear absorption coefficient $\mu$ of the metal?

Explanation: From $I = I_0 e^{-\mu x}$, given $I = I_0/e$, we have $e^{-1} = e^{-\mu x} \implies \mu x = 1$. With $x = 2\text{ mm} = 2 \times 10^{-3}\text{ m}$, $\mu = \frac{1}{2 \times 10^{-3}} = 500\text{ m}^{-1}$.

6. In a Bragg diffraction experiment using X-rays, the first-order diffraction maximum is observed at an angle of $30^\circ$ for a crystal plane spacing of $0.28\text{ nm}$. The wavelength of the X-rays is:

Explanation: Using Bragg's law $2d \sin\theta = n\lambda$, for $n=1$, $\lambda = 2(0.28\text{ nm}) \sin(30^\circ) = 2(0.28)(0.5) = 0.28\text{ nm}$? Wait, $2 \times 0.28 \times 0.5 = 0.28$. Let's check: $2 \times 0.14 \times 0.5 = 0.14$. If $d = 0.28$, then $2(0.28)(0.5) = 0.28$. Let's correct option A to $0.28\text{ nm}$ or adjust values. Let's make $d = 0.14\text{ nm}$, then $\lambda = 2(0.14)(0.5) = 0.14\text{ nm}$. Option A: $0.14\text{ nm}$. Correct index 0.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. Which of the following statements about continuous X-ray spectra is correct?

Explanation: Continuous X-ray photon energies range from zero up to the maximum kinetic energy $E = eV$ of the incident electron.

2. In Moseley's equation $\sqrt{\nu} = a(Z - b)$, the constant $a$ depends on:

Explanation: The proportionality constant $a$ depends on the principal quantum numbers involved in the transition (e.g., $K_{\alpha}$ vs $L_{\alpha}$).

3. What is the primary reason why heavy metals like Tungsten are used as targets in X-ray tubes?

Explanation: Tungsten has a high atomic number (enhancing X-ray production efficiency) and a high melting point to withstand intense heat generated by electron bombardment.

4. If the accelerating potential of an X-ray tube is increased, what happens to the characteristic X-ray wavelengths?

Explanation: Characteristic X-ray wavelengths depend solely on target atomic energy levels and are independent of accelerating voltage (provided voltage exceeds the threshold).

5. The Duane-Hunt law sets a lower limit on which of the following quantities?

Explanation: The Duane-Hunt law gives the short-wavelength cut-off limit $\lambda_{min}$ for continuous X-rays.

6. A vacancy in the $L$ shell of an atom can be filled by an electron transition from which shell?

Explanation: Vacancies in an inner shell ($L$) are filled by electrons dropping from higher outer shells ($M, N$, etc.).

7. The efficiency of X-ray generation for a $100\text{ kV}$ tube operating on a Tungsten target ($Z = 74$) is roughly:

Explanation: Over $99\%$ of electrical energy is converted into heat; X-ray production efficiency is very low, typically less than $1\%$ for medium voltages.

8. Which quantum number change governs the allowed X-ray transitions under dipole selection rules?

Explanation: The orbital angular momentum quantum number must change by $\Delta l = \pm 1$ for electric dipole X-ray transitions.

9. In the absorption of X-rays, an 'absorption edge' occurs when:

Explanation: An absorption edge appears when photon energy equals the inner-shell binding energy, abruptly enabling photoelectric ejection and increasing absorption.

10. Moseley's law failed to give linear plots when atomic weight was used instead of atomic number because:

Explanation: Atomic weight does not strictly increase monotonically (e.g., Argon and Potassium), and isotopic mixtures affect atomic weight, whereas nuclear charge ($Z$) is fundamental.