X-Ray Spectra and Broadening of Spectral Lines
Production of X-rays, and why it is the inverse of the photoelectric effect.
Continuous (bremsstrahlung) spectra and the Duane–Hunt short-wavelength limit.
Characteristic spectra, the \(K\), \(L\), \(M\) series, and Moseley's law.
X-ray absorption and the Auger effect.
Why spectral lines have finite width, and the three broadening mechanisms: natural, Doppler and pressure.
5.1 X-Ray Production
When a target material is bombarded by energetic electrons, X-rays are produced. The process is the inverse of the photoelectric effect: there, a photon is absorbed and an electron ejected; here, an electron is decelerated and a photon emitted.
X-ray wavelengths lie in the range \begin{equation} 0.1\,\mathrm{Å} \lesssim \lambda \lesssim 100\,\mathrm{Å} . \label{eq:xray-range} \end{equation}
5.2 Types of X-Ray Spectra
The X-ray output of such a tube has two quite distinct parts: continuous spectra and characteristic spectra.
5.2.1 Continuous X-ray spectra
This part of the spectrum contains a continuous range of frequencies. It arises from the continuous deceleration of the electrons within the target — the process called bremsstrahlung (“braking radiation”). An accelerated charge radiates, and an electron slowing down in the target field radiates a photon of whatever energy it happens to lose.
The intensity distribution of the continuous spectrum depends on the energy of the incident electrons but is independent of the target material. This contrast with the characteristic spectrum — which depends only on the target — is asked directly.
The most energetic photon possible is emitted when an electron gives up all its kinetic energy \(eV\) in a single collision. This fixes a sharp short-wavelength cutoff: \begin{equation} \boxed{\;\lambda_{\min} = \frac{hc}{E} = \frac{hc}{eV}\;} \label{eq:duane-hunt} \end{equation} where \(V\) is the potential difference through which the electron was accelerated. This is the Duane–Hunt law. In practical units, \begin{equation} \lambda_{\min} = \frac{12400\,\mathrm{eV}\,\mathrm{Å}}{eV} \quad\Longrightarrow\quad \lambda_{\min}\,[\mathrm{Å}] = \frac{12400}{V\,[\mathrm{V}]} . \label{eq:duane-hunt-practical} \end{equation}
Note that \(\lambda_{\min} \propto 1/V\) — so doubling the tube voltage halves the cutoff wavelength. The cutoff depends on nothing but \(V\): not on the target, not on the tube current.
5.2.2 Characteristic X-ray spectra
An energetic electron from the cathode may penetrate deep inside a target atom and knock out an electron from the \(K\) shell. An electron from the \(L\) shell then falls into the vacancy, emitting a photon of energy \begin{equation} h\nu = E_{L} - E_{K} . \label{eq:characteristic} \end{equation} Because atomic energy levels are discrete, the resulting lines are sharp and discrete. They are characteristic of the target element — hence the name.
| Line | Transition | Vacancy in | Series |
|---|---|---|---|
| \(K_\alpha\) | \(L \to K\) | \(K\) (\(n=1\)) | \(K\) series |
| \(K_\beta\) | \(M \to K\) | \(K\) | \(K\) series |
| \(K_\gamma\) | \(N \to K\) | \(K\) | \(K\) series |
| \(L_\alpha\) | \(M \to L\) | \(L\) (\(n=2\)) | \(L\) series |
| \(L_\beta\) | \(N \to L\) | \(L\) | \(L\) series |
5.3 Moseley's Law
The wavenumber of a characteristic line is proportional to the square of the atomic number: \begin{equation} \boxed{\; \tilde\nu = \frac{1}{\lambda} = R_{M}\,(Z-\sigma)^{2} \left(\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}\right)\;} \label{eq:moseley} \end{equation} where
\(Z\) is the atomic number,
\(\sigma\) is the screening constant,
\(R_{M}\) is the Rydberg constant corrected for finite nuclear mass (Chapter 1).
Moseley originally wrote it as \(\sqrt{\nu} = a(Z-b)\), so a plot of \(\sqrt{\nu}\) against \(Z\) is a straight line.
Screening constants worth remembering: \[ K \text{ series}: \ \sigma = 1 , \qquad L \text{ series}: \ \sigma = 7.4 . \] The physical reading of \(\sigma = 1\) for the \(K\) series: the electron making the \(L\to K\) jump sees the nuclear charge \(Ze\) screened by the one remaining \(K\) electron.
Moseley's law established that it is the atomic number, not the atomic weight, that orders the periodic table. It resolved several misplacements in Mendeleev's arrangement and predicted the existence of the then-undiscovered elements \(Z = 43\), \(61\), \(72\) and \(75\).
Calculate the minimum potential that must be applied to produce an X-ray photon of wavelength \(\lambda = 0.1\,\mathrm{Å}\).
Solution. From \(\eqref{eq:duane-hunt-practical}\), \[ \lambda\,[\mathrm{Å}] = \frac{12375}{V\,[\mathrm{V}]} \quad\Longrightarrow\quad V = \frac{12375}{0.1} = 123750\,\mathrm{V} \approx 124\,\mathrm{kV} . \]
The \(K_\alpha\) radiation of molybdenum (\(Z=42\)) has wavelength \(0.72\,\mathrm{Å}\). Calculate the corresponding wavelength for copper (\(Z=29\)).
Solution. For the \(K_\alpha\) line, \(\sigma = 1\), \(n_f = 1\), \(n_i = 2\), so from \(\eqref{eq:moseley}\) \[ \frac{1}{\lambda} = R_{M}(Z-1)^{2}\left(\frac{1}{1^{2}} - \frac{1}{2^{2}}\right) \quad\Longrightarrow\quad \lambda \propto \frac{1}{(Z-1)^{2}} . \] The bracket is the same for both elements and cancels: \[ \frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{Mo}}} = \frac{(42-1)^{2}}{(29-1)^{2}} = \frac{1681}{784} = 2.144 , \] \[ \lambda_{\mathrm{Cu}} = 2.144 \times 0.72 = 1.54\,\mathrm{Å} . \]
This ratio trick — write \(\lambda \propto (Z-\sigma)^{-2}\) and divide — is the fastest route through almost every Moseley question, because the Rydberg constant and the \(n\)-dependent bracket both cancel.
The series limit of the Balmer series of hydrogen is \(364.6\,\mathrm{nm}\). Calculate the atomic number of the element whose \(K\) series extends down to \(0.1\,\mathrm{nm}\).
The source gave this figure as “\(346.6\,\mathrm{cm}\)”. Both the units and the digits are wrong: the Balmer series limit is \(\lambda_\infty = 4/R = 364.6\,\mathrm{nm}\). The final answer \(Z=31\) is unaffected.
Solution. Step 1 — get \(R_M\) from the Balmer limit. For the Balmer series \(n_f = 2\), and the series limit is \(n_i \to \infty\): \[ \frac{1}{\lambda_{\infty}} = R_{M}\left(\frac{1}{2^{2}}\right) \quad\Longrightarrow\quad R_{M} = \frac{4}{\lambda_{\infty}} = \frac{4}{364.6\,\mathrm{nm}} . \]
Step 2 — apply Moseley's law at the \(K\) series limit. For the \(K\) series \(n_f = 1\), \(\sigma = 1\), and the shortest wavelength again corresponds to \(n_i \to \infty\): \[ \frac{1}{\lambda} = R_{M}(Z-1)^{2} \quad\Longrightarrow\quad (Z-1)^{2} = \frac{1}{\lambda R_{M}} = \frac{\lambda_{\infty}}{4\lambda} = \frac{364.6}{4 \times 0.1} = 911.5 . \] Hence \(Z - 1 = 30.2\), giving \[ \boxed{Z \approx 31} \quad (\text{gallium}). \]
5.4 X-Ray Absorption and the Auger Effect
A beam of X-rays passing through matter is attenuated exponentially: \begin{equation} I = I_{0}\,e^{-\mu x} , \label{eq:beer-lambert} \end{equation} where \(\mu\) is the linear absorption coefficient and \(x\) the thickness. The half-value thickness follows at once: \begin{equation} x_{1/2} = \frac{\ln 2}{\mu} . \label{eq:half-value} \end{equation}
5.4.1 The Auger effect
When an inner-shell vacancy is filled, the released energy need not appear as a photon. It can instead be transferred to another bound electron, which is ejected. This is the Auger effect, and the ejected electron is an Auger electron.
If the vacancy is in the \(K\) shell and is filled from \(L\), with a second \(L\) electron ejected, the kinetic energy of the Auger electron is approximately \begin{equation} E_{\text{Auger}} \approx \left(E_{K} - E_{L}\right) - E_{L} = E_{K} - 2E_{L} , \label{eq:auger} \end{equation} writing \(E_{K}\), \(E_{L}\) for the binding energies of the two shells.
Auger emission and X-ray emission are competing de-excitation channels. The fraction going into photons is the fluorescence yield, which rises steeply with \(Z\): Auger emission dominates for light elements, X-ray fluorescence for heavy ones.
In an X-ray tube operating at \(20\,\mathrm{kV}\), calculate the ratio of the de Broglie wavelength of the incident electrons to the shortest wavelength of the generated X-rays.
Solution. The de Broglie wavelength of an electron accelerated through \(V\) is \[ \lambda_{D} = \frac{h}{p} = \frac{h}{\sqrt{2m_{e}eV}} , \] while the shortest X-ray wavelength is \(\lambda_{\min} = hc/eV\) from \(\eqref{eq:duane-hunt}\). Dividing, \[ \frac{\lambda_{D}}{\lambda_{\min}} = \frac{h}{\sqrt{2m_{e}eV}}\cdot\frac{eV}{hc} = \frac{1}{c}\sqrt{\frac{eV}{2m_{e}}} = \frac{1}{c}\sqrt{\frac{V}{2}\cdot\frac{e}{m_{e}}} . \] With \(V = 20\times 10^{3}\,\mathrm{V}\) and \(e/m_{e} = 1.76\times 10^{11}\,\mathrm{C}\,\mathrm{kg}^{-1}\): \[ \frac{\lambda_{D}}{\lambda_{\min}} = \frac{1}{3\times10^{8}} \sqrt{\frac{20\times10^{3}}{2}\times 1.76\times10^{11}} = \frac{1}{3\times10^{8}}\times 4.20\times10^{7} = 0.14 . \]
5.5 Broadening of Spectral Lines
Spectral lines are never perfectly sharp: there is always a finite width. Experimentally this means the wavelength of a line is not precisely defined.
The line width is the separation, in wavenumber (or frequency, or wavelength), between the two points at which the intensity has fallen to half its maximum value — the full width at half maximum (FWHM).
There are three mechanisms:
natural broadening,
Doppler broadening,
collision (pressure) broadening.
5.6 Natural Broadening
This arises from the finite lifetime of energy levels: a level that survives only for a time \(\Delta t\) is not sharply defined in energy. By the energy–time uncertainty relation, \begin{equation} \Delta E\,\Delta t \approx \hbar \quad\Longrightarrow\quad \Delta E \approx \frac{\hbar}{\Delta t} . \label{eq:uncertainty-broadening} \end{equation}
Both the initial and final levels contribute, so the maximum uncertainty in the energy of the emitted photon is \begin{equation} \Delta E = \Delta E_{1} + \Delta E_{2} = \hbar\left(\frac{1}{\Delta t_{1}} + \frac{1}{\Delta t_{2}}\right), \label{eq:natural-energy} \end{equation} and the corresponding frequency width is \begin{equation} \boxed{\; \Delta\nu = \frac{\Delta E}{h} = \frac{1}{2\pi} \left(\frac{1}{\Delta t_{1}} + \frac{1}{\Delta t_{2}}\right)\;} \label{eq:natural-freq} \end{equation}
Watch the factor of \(2\pi\). Since \(\Delta E\,\Delta t \approx \hbar\) (not \(h\)), \[ \Delta\nu = \frac{\Delta E}{h} = \frac{\hbar}{h\,\Delta t} = \frac{1}{2\pi\,\Delta t} , \] which is smaller than the naive \(1/\Delta t\) by a factor of \(2\pi \approx 6.3\). This is the single most common slip in this topic — see the editorial note on JEST 2013 in the solutions.
5.7 Doppler Broadening
The hot atoms that emit spectral lines are not stationary — they move randomly because of thermal agitation. Motion along the line of sight Doppler-shifts the emitted frequency, and since different atoms move at different velocities the line acquires a width.
If the source is at rest relative to the observer, the observed frequency equals the emitted frequency \(\nu_{0}\). Otherwise, for non-relativistic speeds, \begin{align} \text{source approaching:}\quad \nu &= \nu_{0}\left(1 + \frac{V}{c}\right), \label{eq:doppler-toward}\\ \text{source receding:}\quad \nu &= \nu_{0}\left(1 - \frac{V}{c}\right). \label{eq:doppler-away} \end{align}
By the Maxwell–Boltzmann distribution the most probable atomic speed is \begin{equation} V = \sqrt{\frac{2kT}{m}} = \sqrt{\frac{2RT}{M}} , \label{eq:most-probable-speed} \end{equation} where \(m\) is the atomic mass and \(M\) the molar mass. The resulting FWHM is \begin{equation} \boxed{\; \Delta\nu = 1.67\,\frac{\nu_{0}}{c}\sqrt{\frac{2RT}{M}}, \qquad \Delta\lambda = 1.67\,\frac{\lambda_{0}}{c}\sqrt{\frac{2RT}{M}}\;} \label{eq:doppler-width} \end{equation} where \(\nu_{0}\) is the central frequency and \(T\) the absolute temperature.
The factor arises because the FWHM of a Gaussian is \(2\sqrt{\ln 2}\) times the \(1/e\) half-width: \[ 2\sqrt{\ln 2} = 1.6651 . \] The source quoted this as \(1.66\) in one place and \(1.67\) in another; both are roundings of the same number. It has been standardized to \(1.67\) throughout, which is what the source's own worked example uses.
The key scaling is \[ \Delta\nu_{\text{Doppler}} \propto \sqrt{\frac{T}{M}} . \] Hotter gas \(\Rightarrow\) broader; heavier atoms \(\Rightarrow\) narrower. Almost every Doppler-broadening question is this proportionality, and Doppler broadening is normally much larger than natural broadening.
The apparent wavelength of a certain emitted photon is \(5001\,\mathrm{Å}\) whereas its actual wavelength is \(5000\,\mathrm{Å}\). Find the direction and speed of the source relative to the observer.
Solution. Since \(\lambda > \lambda_{0}\) the light is red-shifted, so the source is moving away from the observer. Using \(\eqref{eq:doppler-away}\) with \(\nu = c/\lambda\): \[ \frac{c}{\lambda} = \frac{c}{\lambda_{0}}\left(1 - \frac{V}{c}\right) \quad\Longrightarrow\quad \frac{5000}{5001} = 1 - \frac{V}{c} \quad\Longrightarrow\quad \frac{V}{c} = \frac{1}{5001} . \] Hence \[ V = \frac{3\times10^{8}}{5001} \approx 6.0\times 10^{4}\,\mathrm{m}\,\mathrm{s}^{-1} = 0.6\times10^{5}\ \mathrm{m}\,\mathrm{s}^{-1} . \]
The source's working printed “\(\frac{c}{5001} = \frac{c}{500}\left(1-\frac{V}{c}\right)\)” — the second denominator should be \(5000\), not \(500\). The quoted answer \(0.6\times10^{5}\,\mathrm{m}/\mathrm{s}\) is correct.
Calculate the Doppler half-intensity breadth of the sodium spectral line at \(\lambda_{0} = 5893\,\mathrm{Å}\) at a temperature of \(500\,\mathrm{K}\). Take \(R = 8.31\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}\) and the atomic weight of sodium as \(22.99\,\mathrm{g}\,\mathrm{mol}^{-1}\).
The source's problem statement gave “\(R = 8.31\,\mathrm{erg}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}\)”, but its own solution correctly uses \(R = 8.31\times 10^{7}\,\mathrm{erg}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1} = 8.31\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}\). The statement has been corrected.
Solution. From \(\eqref{eq:doppler-width}\), working in CGS: \[ \Delta\lambda = 1.67\,\frac{\lambda_{0}}{c}\sqrt{\frac{2RT}{M}} = 1.67 \times \frac{5893\times10^{-8}\,\mathrm{cm}}{3\times10^{10}\,\mathrm{cm}\,\mathrm{s}^{-1}} \sqrt{\frac{2\times 8.31\times10^{7}\times 500}{22.99}} . \] Evaluating the square root: \[ \sqrt{\frac{8.31\times10^{10}}{22.99}} = \sqrt{3.615\times10^{9}} = 6.01\times10^{4}\ \mathrm{cm}\,\mathrm{s}^{-1} , \] so \[ \Delta\lambda = 1.67 \times 1.964\times10^{-15} \times 6.01\times10^{4} = 1.97\times10^{-10}\ \mathrm{cm} , \] that is \[ \boxed{\Delta\lambda = 0.0197\,\mathrm{Å}} . \]
5.8 Pressure (Collision) Broadening
As the pressure rises, the distance between molecules shortens, with two consequences:
Collisions between molecules shorten the effective lifetime of the excited state, so by \(\eqref{eq:natural-freq}\) the line width increases.
As molecules approach, their potential fields overlap, which shifts and distorts the levels and changes the natural line width.
The width therefore depends on the number of collisions per second, hence on
the number density of molecules (i.e.\ the pressure), and
the relative speed of the molecules (i.e.\ \(\sqrt{T}\)).
| Mechanism | Physical origin | Depends on | Line shape |
|---|---|---|---|
| Natural | finite level lifetime (uncertainty principle) | lifetime \(\tau\) only | Lorentzian |
| Doppler | thermal motion of emitters | \(\sqrt{T/M}\) | Gaussian |
| Pressure | collisions shorten lifetime; fields overlap | pressure, \(\sqrt{T}\) | Lorentzian |
Natural broadening sets an irreducible floor — it cannot be removed by cooling or by lowering the pressure. Doppler broadening usually dominates in gases at ordinary temperatures, and the observed profile, a convolution of a Gaussian with a Lorentzian, is called a Voigt profile.
5.9 Selection Rules and Polarization
The electric dipole selection rules governing which X-ray transitions occur are the same ones derived in Chapters 3 and 4: for one-electron atoms \(\Delta\ell = \pm1\), \(\Delta j = 0,\pm1\) (but \(j=0 \nrightarrow j=0\)), \(\Delta m_{j} = 0,\pm1\); for many-electron atoms \(\Delta L = 0,\pm1\) (\(L=0\nrightarrow 0\)), \(\Delta J = 0,\pm1\) (\(J=0\nrightarrow0\)), \(\Delta S = 0\); and in all cases the parity of the wavefunction must change.
5.9.1 Polarization of the emitted radiation
Viewed along the positive \(z\) axis (the field direction):
transitions with \(\Delta m_{\ell} = +1\) give left circularly polarized radiation, labelled \(\sigma^{+}\);
transitions with \(\Delta m_{\ell} = -1\) give right circularly polarized radiation, labelled \(\sigma^{-}\);
transitions with \(\Delta m_{\ell} = 0\) (\(\pi\)) are not emitted along \(z\) at all.
The source labelled the \(\Delta m_{\ell} = +1\) component “\(\sigma^{*}\)”. The standard label is \(\sigma^{+}\), paired with \(\sigma^{-}\) for \(\Delta m_{\ell} = -1\).
Formula Summary
X-ray production inverse photoelectric effect; \ \(0.1\,\mathrm{Å} \le \lambda \le 100\,\mathrm{Å}\)
Continuous spectrum (bremsstrahlung) — depends on electron energy, not on target \[ \lambda_{\min} = \frac{hc}{eV} \qquad\text{(Duane--Hunt)}, \qquad \lambda_{\min}[\mathrm{Å}] = \frac{12400}{V[\mathrm{V}]}, \qquad \lambda_{\min}\propto \frac1V \] Characteristic spectrum — depends only on target; \(h\nu = E_L - E_K\) for \(K_\alpha\)
Moseley's law \[ \tilde\nu = \frac{1}{\lambda} = R_M(Z-\sigma)^{2}\left(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}}\right), \qquad \sqrt{\nu} = a(Z-b) \] \[ K \text{ series}: \sigma = 1, \qquad L \text{ series}: \sigma = 7.4, \qquad \lambda \propto (Z-\sigma)^{-2} \] Absorption \(I = I_0 e^{-\mu x}\), \(x_{1/2} = \dfrac{\ln 2}{\mu}\)
Auger effect \(E_{\text{Auger}} \approx E_K - 2E_L\); competes with X-ray emission; fluorescence yield rises with \(Z\)
Line width FWHM — separation of the half-maximum points
Natural broadening \[ \Delta E\,\Delta t \approx \hbar, \qquad \Delta\nu = \frac{1}{2\pi}\left(\frac{1}{\Delta t_1}+\frac{1}{\Delta t_2}\right) \qquad\text{(note the } 2\pi\text{)} \] Doppler broadening \[ \nu = \nu_0\left(1 \pm \frac{V}{c}\right), \qquad V = \sqrt{\frac{2kT}{m}} = \sqrt{\frac{2RT}{M}} \] \[ \Delta\nu = 1.67\frac{\nu_0}{c}\sqrt{\frac{2RT}{M}}, \qquad \Delta\lambda = 1.67\frac{\lambda_0}{c}\sqrt{\frac{2RT}{M}}, \qquad \Delta\nu \propto \sqrt{T/M} \] \(1.67 = 2\sqrt{\ln 2}\), the Gaussian FWHM factor
Pressure broadening \(\propto\) number density (pressure) and \(\sqrt{T}\)
Line shapes natural and pressure \(\to\) Lorentzian; \ Doppler \(\to\) Gaussian; \ convolution \(\to\) Voigt
Polarization along \(\vec B\) \(\Delta m_\ell = +1 \to \sigma^{+}\) (left circular); \ \(\Delta m_\ell = -1 \to \sigma^{-}\) (right circular); \ \(\pi\) not emitted along \(z\)
Previous Year Questions
Group I lists some physical phenomena while Group II gives some parameters. Match each phenomenon with the corresponding parameter.
| Group I | Group II |
|---|---|
| P: Doppler broadening | 1: Moment of inertia |
| Q: Natural broadening | 2: Refractive index |
| R: Rotational spectrum | 3: Lifetime of energy level |
| S: Total internal reflection | 4: Pressure |
- (a)
P-4, Q-3, R-1, S-2
- (b)
P-3, Q-2, R-1, S-4
- (c)
P-2, Q-3, R-4, S-1
- (d)
P-1, Q-4, R-2, S-3
The target of an X-ray tube is subjected to an excitation voltage \(V\). The wavelength of the X-rays produced is proportional to
- (a)
\(\dfrac{1}{V}\)
- (b)
\(\sqrt{\dfrac{1}{V}}\)
- (c)
\(\sqrt{V}\)
- (d)
\(V\)
X-rays are produced using cobalt (\(Z=27\)) as target. The spectrum contains a strong \(K_\alpha\) line of wavelength \(0.1785\,\mathrm{nm}\) and a weak \(K_\alpha\) line of wavelength \(0.1930\,\mathrm{nm}\). The weak \(K_\alpha\) line is due to an impurity whose atomic number is
- (a)
\(25\)
- (b)
\(26\)
- (c)
\(28\)
- (d)
\(30\)
A beam of X-rays of intensity \(I_{0}\) is incident normally on a metal sheet of thickness \(2\,\mathrm{mm}\). The intensity of the transmitted beam is \(0.025\,I_{0}\). The absorption coefficient of the metal sheet, in \(\mathrm{m}^{-1}\), is .
The curves P and Q schematically show the variation of X-ray intensity with wavelength at two different accelerating voltages, for a given target material. In the figure \(\lambda_{1} = 0.25\,\mathrm{Å}\), \(\lambda_{2} = 0.5\,\mathrm{Å}\), \(\lambda_{3} = 1.0\,\mathrm{Å}\) and \(\lambda_{4} = 2.25\,\mathrm{Å}\). Take Planck's constant as \(6.6\times 10^{-34}\,\mathrm{J}\,\mathrm{s}\), the speed of light as \(3\times 10^{8}\,\mathrm{m}\,\mathrm{s}^{-1}\) and the elementary charge as \(1.6\times 10^{-19}\,\mathrm{C}\).
Which of the following statement(s) is/are true?
- (a)
The accelerating potential corresponding to curve P is greater than that of curve Q
- (b)
The accelerating potential applied to obtain curve Q is \(24750\,\mathrm{V}\)
- (c)
Peaks (II) and (IV) correspond to radiative transitions from the \(L\) to the \(K\) shell
- (d)
Peaks (I) and (III) correspond to radiative transitions from the \(N\) to the \(K\) shell
Diffuse hydrogen gas within a galaxy may be assumed to follow a Maxwell distribution at temperature \(e6\,\mathrm{K}\), while the temperature appropriate for the H gas in intergalactic space, following the same distribution, may be taken as \(e4\,\mathrm{K}\). The ratio of thermal broadening \(\Delta\nu_{G}/\Delta\nu_{\mathrm{IG}}\) of the Lyman-\(\alpha\) line from H atoms within the galaxy to that from intergalactic space is closest to
- (a)
\(100\)
- (b)
\(\dfrac{1}{100}\)
- (c)
\(10\)
- (d)
\(\dfrac{1}{10}\)
Argon (\(A = 40\)) emits a line at \(\lambda = 550\,\mathrm{nm}\) at a temperature \(T = 400\,\mathrm{K}\). The full Doppler width \(\Delta\lambda\) is closest to
- (a)
\(10^{-2}\ \mathrm{nm}\)
- (b)
\(10^{-1}\ \mathrm{nm}\)
- (c)
\(10^{-3}\ \mathrm{nm}\)
- (d)
\(10^{-5}\ \mathrm{nm}\)
A sodium atom in the first excited \(3P\) state has a lifetime of \(16\,\mathrm{ns}\) for decaying to the ground \(3S\) state. The wavelength of the emitted photon is \(589\,\mathrm{nm}\). The corresponding line width of the transition, in frequency units, is about
- (a)
\(1.7\times 10^{6}\,\mathrm{Hz}\)
- (b)
\(1\times 10^{7}\,\mathrm{Hz}\)
- (c)
\(6.3\times 10^{7}\,\mathrm{Hz}\)
- (d)
\(5\times 10^{14}\,\mathrm{Hz}\)
If a hydrogen atom is bombarded by energetic electrons, it will emit
- (a)
\(K_\alpha\) X-rays
- (b)
\(\beta\) rays
- (c)
neutrons
- (d)
none of the above
Solutions to Previous Year Questions
Solution. Take the three unambiguous pairings first:
Q: Natural broadening \(\to\) 3 (lifetime of energy level), from \(\Delta E\,\Delta t \approx \hbar\).
R: Rotational spectrum \(\to\) 1 (moment of inertia), since the rotational constant is \(B = \hbar^{2}/2I\) (Chapter 7).
S: Total internal reflection \(\to\) 2 (refractive index), via \(\sin\theta_{c} = n_{2}/n_{1}\).
By elimination P: Doppler broadening \(\to\) 4 (pressure), giving P-4, Q-3, R-1, S-2.
Strictly, Doppler broadening is governed by temperature, not pressure — pressure is what governs collision broadening. The matching is only forced because “temperature” is not among the options and the other three pairings are unambiguous. Worth pointing out to students so they answer by elimination rather than trying to justify P-4 physically.
Solution. From the Duane–Hunt law \(\eqref{eq:duane-hunt}\), \[ \lambda_{\min} = \frac{hc}{eV} \propto \frac{1}{V} . \]
Solution. For \(K_\alpha\) lines, Moseley's law gives \(\dfrac{1}{\lambda} \propto (Z-1)^{2}\). Writing this for the two lines, \[ \frac{1}{0.1785} \propto (27-1)^{2}, \qquad \frac{1}{0.1930} \propto (Z-1)^{2} . \] Dividing the second by the first, \[ \frac{(Z-1)^{2}}{26^{2}} = \frac{0.1785}{0.1930} = 0.9249 , \] \[ (Z-1)^{2} = 676 \times 0.9249 = 625.2 \quad\Longrightarrow\quad Z - 1 = 25.0 \quad\Longrightarrow\quad \boxed{Z = 26} \ \ (\text{iron}). \]
The source's working printed the first relation as “\(\frac{1}{0.18750} \propto (27-1)^{2}\)”. The wavelength given in the question is \(0.1785\,\mathrm{nm}\); the digits \(0.18750\) are a transcription slip. The final answer \(Z=26\) is unaffected.
Solution. From the Beer–Lambert law (6.13) with \(x = 2\times 10^{-3}\,\mathrm{m}\): \[ 0.025\,I_{0} = I_{0}\,e^{-\mu x} \quad\Longrightarrow\quad \mu = \frac{-\ln(0.025)}{2\times10^{-3}} = \frac{3.689}{2\times10^{-3}} = 1.84\times 10^{3}\,\mathrm{m}^{-1} . \]
Solution. Two facts settle all four options: the short-wavelength cut-off of the continuum depends only on the accelerating voltage, and the characteristic peaks depend only on the target.
(a) True. Curve P is cut off at \(\lambda_{1} = 0.25\,\mathrm{Å}\) and curve Q at \(\lambda_{2} = 0.5\,\mathrm{Å}\). By the Duane–Hunt law \(\eqref{eq:duane-hunt}\), \(\lambda_{\min} = hc/eV\), so a shorter cut-off means a larger voltage: \(V_{P} > V_{Q}\).
(b) True. For curve Q, \[ V_{Q} = \frac{hc}{e\lambda_{2}} = \frac{6.6\times10^{-34}\times3\times10^{8}} {1.6\times10^{-19}\times0.5\times10^{-10}} = \frac{1.98\times10^{-25}}{8\times10^{-30}} = 24750\,\mathrm{V}. \]
(c) True. Both curves show their characteristic peaks at the same two wavelengths, as they must for a single target. The pair at \(\lambda_{4} = 2.25\,\mathrm{Å}\) — peaks (II) and (IV) — is the longer-wavelength, hence lower-energy, member of the \(K\) series: that is \(K_\alpha\), the \(L \to K\) transition.
(d) False. Peaks (I) and (III) at \(\lambda_{3} = 1.0\,\mathrm{Å}\) are the higher-energy member, \(K_\beta\), which is the \(M \to K\) transition, not \(N \to K\).
Keep the working form of the Duane–Hunt law to hand: \(\lambda_{\min}[\mathrm{Å}] = 12400/V[\mathrm{V}]\). Here \(12400/0.5 = 24800\,\mathrm{V}\), matching (b) to within the rounding of \(hc\).
The single idea behind (a), (c) and (d) together is that raising the voltage moves the continuum cut-off but leaves the characteristic lines exactly where they were. That is the whole content of Moseley's law: the line wavelengths are a property of the atom, not of the tube.
Solution. “Thermal” broadening means Doppler broadening, for which \(\eqref{eq:doppler-width}\) gives \[ \Delta\nu_{D} = \frac{\nu_{0}}{c}\sqrt{\frac{2kT}{m}} \quad\Longrightarrow\quad \Delta\nu_{D} \propto \sqrt{T} . \] The same line (\(\text{Lyman-}\alpha\)) from the same species (H) means \(\nu_{0}\) and \(m\) cancel, so \[ \frac{\Delta\nu_{G}}{\Delta\nu_{\mathrm{IG}}} = \sqrt{\frac{T_{G}}{T_{\mathrm{IG}}}} = \sqrt{\frac{10^{6}}{10^{4}}} = \sqrt{100} = 10 . \]
Solution. The full width at half maximum of a Doppler-broadened line is \[ \Delta\lambda = \frac{2\lambda}{c}\sqrt{\frac{2k_{B}T\ln2}{M}} . \] The mass of an argon atom is \(M = 40\times1.67\times10^{-27} = 6.68\times 10^{-26}\,\mathrm{kg}\), so \[ \frac{2k_{B}T\ln2}{M} = \frac{2\times1.38\times10^{-23}\times400\times0.693}{6.68\times10^{-26}} = \frac{7.65\times10^{-21}}{6.68\times10^{-26}} = 1.15\times10^{5}\ \mathrm{m}^{2}\,\mathrm{s}^{-2}, \] whose square root is \(338\,\mathrm{m}\,\mathrm{s}^{-1}\). Hence \[ \Delta\lambda = \frac{2\times550\times10^{-9}}{3\times10^{8}}\times338 = 1.24\times10^{-12}\ \mathrm{m} = 1.24\times10^{-3}\ \mathrm{nm} \approx \boxed{10^{-3}\ \mathrm{nm}} . \]
An order-of-magnitude route gets there without touching the constant: \(\Delta\lambda/\lambda \approx v_{\text{th}}/c\), and a light atom at room temperature moves at a few hundred metres per second, so \(\Delta\lambda/\lambda \approx 300/(3\times10^{8}) \approx 10^{-6}\). Then \(\Delta\lambda \approx 550\times10^{-6}\ \mathrm{nm} \approx 10^{-3}\ \mathrm{nm}\) to the nearest decade. On a question that asks only for the decade, do this and move on.
Solution. Natural broadening from a single finite lifetime \(\tau = 16\,\mathrm{ns}\). Using \(\Delta E\,\Delta t \approx \hbar\) as in \(\eqref{eq:natural-freq}\), \[ \Delta\nu = \frac{\Delta E}{h} = \frac{\hbar}{h\tau} = \frac{1}{2\pi\tau} = \frac{1}{2\pi \times 16\times10^{-9}} = 9.95\times 10^{6}\,\mathrm{Hz} \approx 1.0\times 10^{7}\,\mathrm{Hz} . \]
The source's solution wrote \(\Delta E\cdot\Delta t = \hbar\) and then evaluated \(\Delta f = 1/\Delta t = 6.25\times 10^{7}\,\mathrm{Hz}\), selecting option (c). Those two steps are inconsistent: \(\Delta E\,\Delta t = \hbar\) gives \(\Delta\nu = 1/(2\pi\tau)\), not \(1/\tau\). The two readings are: \[ \Delta E\,\Delta t = \hbar \ \Rightarrow\ \Delta\nu = \frac{1}{2\pi\tau} = 9.95\times 10^{6}\,\mathrm{Hz} \ \to \text{option (b)} , \] \[ \Delta E\,\Delta t = h \ \Rightarrow\ \Delta\nu = \frac{1}{\tau} = 6.25\times 10^{7}\,\mathrm{Hz} \ \to \text{option (c)} . \] Note also that both \(1\times 10^{7}\,\mathrm{Hz}\) and \(6.3\times 10^{7}\,\mathrm{Hz}\) are offered as options, which strongly suggests the paper intended to discriminate between exactly these two conventions. The physically standard natural linewidth is \(\Delta\nu = 1/(2\pi\tau)\), so (b) is given above — but this contradicts the source's key, and the official JEST answer should be checked.
Solution. Hydrogen has only one electron. Knocking it out simply ionizes the atom; there is no outer-shell electron available to fall into the vacancy, so no \(K_\alpha\) line can be produced. \(\beta\) rays come from nuclear decay and neutrons from nuclear reactions, neither of which electron bombardment induces. Hence none of the above.
Characteristic X-rays require at least two occupied shells. This is why \(K_\alpha\) lines only exist for \(Z \ge 3\) (lithium onwards) — another way the same point gets asked.