IPhO 1988 · Theory problem 1

Spectroscopy of Particle Velocities

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Basic Data

The absorption and emission of a photon is a reversible process. A good example is to be found in the excitation of an atom from the ground state to a higher energy state and the atoms´ subsequent return to the ground state. In such a case we may detect the absorption of a photon from the phenomenon of spontaneous emission or fluorescence. Some of the more modern instrumentation make use of this principle to identify atoms, and also to measure or calculate the value of the velocity in the velocity spectrum of the electron beam.

In an idealised experiment (see fig. 19.1) a single-charged ion travels in the opposite direction to light from a laser source with velocity \(v\). The wavelength of light from the laser source is adjustable. An ion with velocity Zero can be excited to a higher energy state by the application of laser light having a wavelength of \(\lambda=600\ \mathrm{nm}\). If we excite a moving ion, our knowledge on Dopplers´ effect tells us that we need to apply laser light of a wavelength other than the value given above.

There is given a velocity spectrum embracing velocity magnitude from \(v_1=0\ \frac{\mathrm m}{\mathrm s}\) to \(v_2=6{,}000\ \frac{\mathrm m}{\mathrm s}\). (see fig. 19.1)

Figure 1

Fig. 19.1

Questions

1.1

1.1.1

What range of wavelength of the laser beam must be used to excite ions of all velocities in the velocity spectrum given above ?

1.1.2

A rigorous analysis of the problem calls for application of the principle from the theory of special relativity

\[ \nu'=\nu\cdot\sqrt{\frac{1+\dfrac vc}{1-\dfrac vc}} \]

Determine the error when the classical formula for Dopplers´ effect is used to solve the problem.

1.2

Assuming the ions are accelerated by a potential \(U\) before excited by the laser beam, determine the relationship between the width of the velocity spectrum of the ion beam and the accelerating potential. Does the accelerating voltage increase or decrease the velocity spectrum width ?

1.3

Each ion has the value \(\dfrac em=4\cdot10^6\ \dfrac{\mathrm{A\cdot s}}{\mathrm{kg}}\), two energy levels corresponding to wavelength \(\lambda^{(1)}=600\ \mathrm{nm}\) and wavelength \(\lambda^{(2)}=\lambda^{(1)}+10^{-3}\ \mathrm{nm}\). Show that lights of the two wavelengths used to excite ions overlap when no accelerating potential is applied. Can accelerating voltage be used to separate the two spectra of laser light used to excite ions so that they no longer overlap ? If the answer is positive, calculate the minimum value of the voltage required.

Figure 2

Fig. 19.2

Official English text of the problem.

Answers

  • 1.1.1: The required laser wavelength range is \[ \boxed{600.000\ \mathrm{nm}\leq\lambda_{\mathrm L}\leq600.012\ \mathrm{nm}}, \] with width \(0.012\ \mathrm{nm}\).

  • 1.1.2: The classical formula underestimates the upper wavelength by approximately \[ \boxed{1.2\times10^{-7}\ \mathrm{nm}}. \] The maximum relative error in wavelength is \(2.0\times10^{-10}\); the relative error in the scan width is \(1.0\times10^{-5}\).

  • 1.2: After acceleration, the velocity-spectrum width is \[ \boxed{\Delta v(U)= \sqrt{v_2^2+\frac{2eU}{m}} -\sqrt{v_1^2+\frac{2eU}{m}}}. \] It decreases as the accelerating voltage increases.

  • 1.3: Without acceleration, the two wavelength intervals are approximately \[ [600.000,600.012]\ \mathrm{nm}, \qquad [600.001,600.013]\ \mathrm{nm}, \] so they overlap. Acceleration can separate them, with threshold \[ \boxed{U_{\min}\approx160\ \mathrm V}. \] At the threshold the bands just touch; a slightly larger voltage produces a finite gap.

Solution

Use \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\). Let \(\lambda_0\) denote an ion’s resonance wavelength at rest, and let \(\lambda_{\mathrm L}\) denote the laser wavelength in the laboratory. The idealized spectra are taken to have no intrinsic linewidth.

1.1.1

The ion moves towards the incoming light, so it observes a higher frequency than the laboratory laser frequency. In the classical moving-observer approximation,

\[ \nu'=\nu_{\mathrm L}\left(1+\frac vc\right). \]

Excitation requires the frequency seen by the ion to equal its rest resonance frequency:

\[ \nu'=\nu_0=\frac{c}{\lambda_0}. \]

Since \(\nu_{\mathrm L}=c/\lambda_{\mathrm L}\),

\[ \frac{c}{\lambda_0} =\frac{c}{\lambda_{\mathrm L}}\left(1+\frac vc\right), \]

and therefore

\[ \boxed{\lambda_{\mathrm L}(v)=\lambda_0\left(1+\frac vc\right)}. \]

Thus the laboratory laser wavelength must increase with ion speed. At the endpoints,

\[ \lambda_{\mathrm L}(0)=600\ \mathrm{nm}, \]

\[ \lambda_{\mathrm L}(6000) =600\left(1+\frac{6000}{3.00\times10^8}\right)\mathrm{nm} =600.012\ \mathrm{nm}. \]

The required scan width is consequently

\[ \boxed{\Delta\lambda_{\mathrm L} =\frac{\lambda_0(v_2-v_1)}{c} =0.012\ \mathrm{nm}}. \]

1.1.2

Put \(\beta=v/c\). The supplied relativistic formula gives the resonance condition

\[ \nu_0=\nu_{\mathrm L}\sqrt{\frac{1+\beta}{1-\beta}}, \]

so the exact required wavelength is

\[ \lambda_{\mathrm{rel}} =\lambda_0\sqrt{\frac{1+\beta}{1-\beta}} =\lambda_0\frac{1+\beta}{\sqrt{1-\beta^2}}. \]

The classical value is \(\lambda_{\mathrm{cl}}=\lambda_0(1+\beta)\). Hence its fractional underestimate, measured relative to the exact wavelength, is

\[ \epsilon_\lambda =\frac{\lambda_{\mathrm{rel}}-\lambda_{\mathrm{cl}}} {\lambda_{\mathrm{rel}}} =1-\sqrt{1-\beta^2} \simeq\frac{\beta^2}{2}. \]

This error is greatest at \(v_2=6000\ \mathrm{m\,s^{-1}}\), for which

\[ \beta_2=2.00\times10^{-5}, \qquad \boxed{\epsilon_{\lambda,\max}\simeq2.00\times10^{-10}}. \]

Equivalently, expanding the exact wavelength,

\[ \lambda_{\mathrm{rel}} =\lambda_0\left(1+\beta+\frac{\beta^2}{2} +O(\beta^3)\right), \]

gives the maximum absolute correction

\[ \delta\lambda_{\max} \simeq\frac{\lambda_0\beta_2^2}{2} =600\ \mathrm{nm}\times2.00\times10^{-10} =\boxed{1.20\times10^{-7}\ \mathrm{nm}}. \]

Thus the relativistic upper endpoint is approximately

\[ 600.01200012\ \mathrm{nm}. \]

The lower endpoint has no correction because \(v_1=0\). Therefore the absolute error in the scan width is also \(1.20\times10^{-7}\ \mathrm{nm}\), corresponding to

\[ \frac{\delta(\Delta\lambda_{\mathrm L})}{\Delta\lambda_{\mathrm L}} \simeq\frac{1.20\times10^{-7}}{0.012} =\boxed{1.00\times10^{-5}} \]

or \(0.001\%\).

1.2

Let \(u\) be an ion’s speed after acceleration. Every singly charged ion gains the same kinetic energy \(eU\). Using nonrelativistic energy conservation,

\[ \frac12mu^2=\frac12mv^2+eU, \]

so

\[ u(v,U)=\sqrt{v^2+\frac{2eU}{m}}. \]

This function increases with the initial speed \(v\), so the endpoints of the accelerated distribution are

\[ u_1=\sqrt{v_1^2+\frac{2eU}{m}}, \qquad u_2=\sqrt{v_2^2+\frac{2eU}{m}}. \]

Consequently,

\[ \boxed{\Delta v(U)=u_2-u_1 =\sqrt{v_2^2+\frac{2eU}{m}} -\sqrt{v_1^2+\frac{2eU}{m}}}. \]

Rationalizing makes its dependence on voltage especially clear:

\[ \Delta v(U)= \frac{v_2^2-v_1^2} {\sqrt{v_2^2+2eU/m}+\sqrt{v_1^2+2eU/m}}. \]

The numerator is constant and the denominator increases with \(U\). Thus acceleration shifts the whole distribution towards higher speeds while decreasing its width.

For the given \(v_1=0\),

\[ \boxed{\Delta v(U)= \sqrt{v_2^2+\frac{2eU}{m}} -\sqrt{\frac{2eU}{m}}}. \]

When \(2eU/m\gg v_2^2\), this becomes

\[ \Delta v(U)\simeq \frac{v_2^2}{2\sqrt{2eU/m}} \propto U^{-1/2}. \]

1.3

Write

\[ \lambda_1=600\ \mathrm{nm},\qquad \lambda_2=600.001\ \mathrm{nm},\qquad \delta\lambda=\lambda_2-\lambda_1=0.001\ \mathrm{nm}. \]

Without acceleration, each rest wavelength produces a laser excitation interval

\[ \mathcal I_i(0)= \left[\lambda_i,\lambda_i\left(1+\frac{v_2}{c}\right)\right]. \]

Numerically,

\[ \mathcal I_1(0)=[600.000,600.012]\ \mathrm{nm}, \]

\[ \mathcal I_2(0)\simeq[600.001,600.013]\ \mathrm{nm}. \]

Their common interval is approximately

\[ [600.001,600.012]\ \mathrm{nm}, \]

so the two spectra clearly overlap.

After acceleration, the intervals become

\[ \mathcal I_i(U)= \left[ \lambda_i\left(1+\frac{u_1}{c}\right), \lambda_i\left(1+\frac{u_2}{c}\right) \right]. \]

Since \(\lambda_2>\lambda_1\), separation occurs when the upper edge of the first interval lies below the lower edge of the second. The limiting condition is

\[ \lambda_1\left(1+\frac{u_2}{c}\right) =\lambda_2\left(1+\frac{u_1}{c}\right). \]

Equivalently,

\[ u_2-u_1 =\frac{c\,\delta\lambda}{\lambda_1} \left(1+\frac{u_1}{c}\right). \]

To leading Doppler order, \(u_1/c\ll1\), so the required limiting velocity width is

\[ d\equiv\frac{c\,\delta\lambda}{\lambda_1} =\frac{(3.00\times10^8)(0.001)}{600}\ \mathrm{m\,s^{-1}} =\boxed{500\ \mathrm{m\,s^{-1}}}. \]

Acceleration can achieve this because it continuously narrows the velocity distribution.

With \(v_1=0\), energy conservation also gives

\[ u_2^2-u_1^2=v_2^2. \]

At the separation threshold, \(u_2-u_1\simeq d\), and therefore

\[ u_2+u_1\simeq\frac{v_2^2}{d} =\frac{6000^2}{500} =72{,}000\ \mathrm{m\,s^{-1}}. \]

Solving for the two speeds,

\[ u_1\simeq\frac{72{,}000-500}{2} =35{,}750\ \mathrm{m\,s^{-1}}, \]

\[ u_2\simeq36{,}250\ \mathrm{m\,s^{-1}}. \]

Finally, using \(e/m=4\times10^6\ \mathrm{C\,kg^{-1}}\),

\[ U_{\min}\simeq\frac{u_1^2}{2(e/m)} =\frac{(35{,}750)^2}{8\times10^6}\ \mathrm V \approx\boxed{160\ \mathrm V}. \]

The resulting speeds satisfy \(u/c\approx1.2\times10^{-4}\), consistent with the approximations used. Retaining the small factor \(1+u_1/c\) in the classical endpoint condition does not change the reported voltage.

At the limiting voltage the two spectral edges coincide. Strict separation requires a voltage slightly above this threshold.

The solution was checked against the official solution.