Week 12: Atomic Spectra, X-Rays & Nuclear Physics

\(\Delta\) ← if you see the word "Delta" instead of a triangle, please refresh the page! Thanks :)

[1] Overview⚓︎

I recommend reviewing the following concepts, they will probably come up in recitation:

• Hydrogen energy levels and photon transitions

  • \(E_n = -\dfrac{13.6\ \mathrm{eV}}{n^2}\)
  • knowing when a transition is emission vs. absorption
  • finding photon energy from energy-level differences

• Spectral series

  • Lyman series (\(n_f=1\))
  • Balmer series (\(n_f=2\))
  • Paschen series (\(n_f=3\))
  • connecting series names to wavelength / energy trends

• Hydrogen-like ions

  • \(E_n = -\dfrac{13.6Z^2\ \mathrm{eV}}{n^2}\)
  • examples like \(\mathrm{He^+}\) and \(\mathrm{Li^{2+}}\)
  • how increasing \(Z\) changes level spacing and photon energy

• Total energy, kinetic energy, and potential energy in bound atomic states

  • \(E = K + U\)
  • for Coulomb orbits, \(K=-E\) and \(U=2E\)
  • ionization energy from an excited state

• X-ray spectra from x-ray tubes

  • background (bremsstrahlung) radiation
  • the cutoff wavelength \(\lambda_0\)
  • what changing the voltage does
  • what changing the target material does

• Nuclear composition, mass defect, and binding energy

  • counting protons and neutrons
  • using atomic mass data carefully
  • mass defect \(\Delta m\)
  • binding energy and binding energy per nucleon

Practice Problems⚓︎

• If you're comfortable with the "★★★" problems, you should do great during recitation.
• To print these questions, simply press Ctrl + P while on this page, and it should come out formatted nicely -- just make sure to refresh first so all the math renders properly.
  • For a copy without solutions, click here.

Difficulty key:

• ★☆☆ = beginner
• ★★☆ = standard
• ★★★ = challenging / multi-step

[2.1] Hydrogen Energy Levels and Spectral Lines⚓︎

1. ★☆☆
   A hydrogen atom starts in the ground state and absorbs a photon that excites the electron to the \(n=3\) energy level.

   • (a) Is this an emission or absorption process?
   • (b) Find the photon energy in eV.
   • (c) Find the wavelength of the photon.
   Solution

   For hydrogen,

   \[E_n = -\frac{13.6\ \mathrm{eV}}{n^2}.\]

   The initial and final energies are

   \[E_1 = -13.6\ \mathrm{eV}, \qquad E_3 = -\frac{13.6}{9} = -1.51\ \mathrm{eV}.\]

   Since the atom goes to a higher energy level, this is an absorption process.

   The photon energy must equal the increase in atomic energy:

   \[E_\gamma = E_3 - E_1 = (-1.51) - (-13.6) = 12.09\ \mathrm{eV}.\]

   The wavelength is

   \[\lambda = \frac{hc}{E_\gamma} = \frac{1240\ \mathrm{eV \cdot nm}}{12.09\ \mathrm{eV}} = 102.6\ \mathrm{nm}.\]

   
   Answer: (a) absorption; (b) \(12.09\ \mathrm{eV}\); (c) \(\lambda \approx 102.6\ \mathrm{nm}\).

2. ★★☆
   In a hydrogen atom, an electron makes a transition from \(n=6\) to \(n=3\).

   • (a) Is a photon emitted or absorbed?
   • (b) Find the energy of the photon.
   • (c) Find the wavelength of the photon.
   • (d) Identify the spectral series.
   Solution

   Because the electron moves from a higher level to a lower level, a photon is emitted.

   The energy difference is

   \[E_\gamma = 13.6\left(\frac{1}{3^2}-\frac{1}{6^2}\right)\mathrm{eV}.\]

   So

   \[E_\gamma = 13.6\left(\frac{1}{9}-\frac{1}{36}\right)=13.6\left(\frac{3}{36}\right)=1.13\ \mathrm{eV}.\]

   Then

   \[\lambda = \frac{1240}{1.13} \approx 1.09 \times 10^3\ \mathrm{nm}.\]

   More precisely,

   \[\lambda \approx 1094\ \mathrm{nm}.\]

   Since the final level is \(n_f=3\), this is in the Paschen series.

   
   Answer: (a) emitted; (b) \(1.13\ \mathrm{eV}\); (c) \(\lambda \approx 1094\ \mathrm{nm}\); (d) Paschen series.

3. ★★★
   A hydrogen electron starts at \(n=4\) and de-excites in two steps: first \(n=4 \to 2\), then \(n=2 \to 1\).

   • (a) Find the energy and wavelength of each emitted photon.
   • (b) Find the total energy emitted in the two-step process.
   • (c) Compare this total to the energy of the single direct transition \(n=4 \to 1\).
   Solution

   For the first photon,

   \[E_{\gamma,1} = 13.6\left(\frac{1}{2^2}-\frac{1}{4^2}\right) = 13.6\left(\frac{1}{4}-\frac{1}{16}\right)=13.6\left(\frac{3}{16}\right)=2.55\ \mathrm{eV}.\]

   Its wavelength is

   \[\lambda_1 = \frac{1240}{2.55} = 486\ \mathrm{nm}.\]

   For the second photon,

   \[E_{\gamma,2} = 13.6\left(1-\frac{1}{4}\right)=13.6\left(\frac{3}{4}\right)=10.2\ \mathrm{eV}.\]

   Its wavelength is

   \[\lambda_2 = \frac{1240}{10.2} = 121.6\ \mathrm{nm}.\]

   The total emitted energy is

   \[E_{\mathrm{total}} = E_{\gamma,1}+E_{\gamma,2}=2.55+10.2=12.75\ \mathrm{eV}.\]

   Now compare with the direct transition \(n=4 \to 1\):

   \[E_{4\to1}=13.6\left(1-\frac{1}{16}\right)=13.6\left(\frac{15}{16}\right)=12.75\ \mathrm{eV}.\]

   So the two-step process emits the same total energy as the direct transition, just split between two photons instead of one.

   
   Answer: (a) \(4\to2\): \(2.55\ \mathrm{eV}\) and \(486\ \mathrm{nm}\); \(2\to1\): \(10.2\ \mathrm{eV}\) and \(121.6\ \mathrm{nm}\); (b) \(12.75\ \mathrm{eV}\) total; (c) this equals the energy of the direct \(4\to1\) transition.

4. ★★★
   One hydrogen atom is excited to the \(n=5\) state.

   • (a) What is the maximum possible number of photons this atom can emit before it reaches the ground state?
   • (b) If a large collection of hydrogen atoms is prepared in the \(n=5\) state, how many different photon wavelengths could appear in the emitted spectrum as the atoms relax?
   • (c) Among all of those possible emitted photons, which transition gives the longest wavelength, and which gives the shortest wavelength? Find both wavelengths.
   Solution

   The maximum number of photons from a single atom occurs if it drops only one level at a time:

   \[5 \to 4 \to 3 \to 2 \to 1.\]

   That gives

   \[4 \text{ photons}.\]

   For a large collection of atoms initially in \(n=5\), different atoms can follow different downward paths. The possible downward transitions among the levels \(1,2,3,4,5\) are

   \[5\to4,\ 5\to3,\ 5\to2,\ 5\to1,\ 4\to3,\ 4\to2,\ 4\to1,\ 3\to2,\ 3\to1,\ 2\to1.\]

   So the number of distinct emitted wavelengths is

   \[\binom{5}{2}=10.\]

   The longest wavelength corresponds to the smallest energy difference, which is the transition \(5\to4\):

   \[E_{5\to4}=13.6\left(\frac{1}{4^2}-\frac{1}{5^2}\right) =13.6\left(\frac{1}{16}-\frac{1}{25}\right) =13.6\left(\frac{9}{400}\right)=0.306\ \mathrm{eV}.\]

   Thus,

   \[\lambda_{\mathrm{long}}=\frac{1240}{0.306}=4.05\times10^3\ \mathrm{nm}.\]

   So the longest wavelength is about

   \[\lambda_{\mathrm{long}}\approx 4050\ \mathrm{nm}.\]

   The shortest wavelength corresponds to the largest energy difference, which is the transition \(5\to1\):

   \[E_{5\to1}=13.6\left(1-\frac{1}{25}\right)=13.6\left(\frac{24}{25}\right)=13.06\ \mathrm{eV}.\]

   Therefore,

   \[\lambda_{\mathrm{short}}=\frac{1240}{13.06}=94.9\ \mathrm{nm}.\]

   
   Answer: (a) maximum of 4 photons; (b) 10 different wavelengths; (c) longest wavelength from \(5\to4\), about \(4050\ \mathrm{nm}\), and shortest wavelength from \(5\to1\), about \(94.9\ \mathrm{nm}\).

[2.2] Hydrogen-Like Ions, Total Energy, and Ionization⚓︎

1. ★☆☆
   Find the total energy of an electron in the \(n=4\) state of the hydrogen-like ion \(\mathrm{He^+}\).

   Solution

   For a hydrogen-like ion,

   \[E_n = -\frac{13.6Z^2}{n^2}\ \mathrm{eV}.\]

   For \(\mathrm{He^+}\), \(Z=2\), so

   \[E_4 = -\frac{13.6(2^2)}{4^2} = -\frac{54.4}{16} = -3.40\ \mathrm{eV}.\]

   
   Answer: \(E_4 = -3.40\ \mathrm{eV}\).

2. ★★☆
   An electron in the hydrogen-like ion \(\mathrm{Li^{2+}}\) makes a transition from \(n=5\) to \(n=2\).

   • (a) Find the energy of the emitted photon.
   • (b) Find the wavelength of the photon.
   Solution

   For \(\mathrm{Li^{2+}}\), the atomic number is \(Z=3\).

   The photon energy is

   \[E_\gamma = 13.6Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right).\]

   So

   \[E_\gamma = 13.6(3^2)\left(\frac{1}{2^2}-\frac{1}{5^2}\right)\]

   \[E_\gamma = 122.4\left(\frac{1}{4}-\frac{1}{25}\right) = 122.4\left(\frac{21}{100}\right)=25.7\ \mathrm{eV}.\]

   Then

   \[\lambda = \frac{1240}{25.7} = 48.2\ \mathrm{nm}.\]

   
   Answer: (a) \(25.7\ \mathrm{eV}\); (b) \(\lambda \approx 48.2\ \mathrm{nm}\).

3. ★★★
   An electron in \(\mathrm{He^+}\) is in the \(n=3\) state.

   • (a) Find its total energy.
   • (b) Find how much of this energy is kinetic energy.
   • (c) Find how much of this energy is electric potential energy.
   • (d) Find the minimum photon energy required to ionize the electron from this state.
   Solution

   For \(\mathrm{He^+}\), \(Z=2\), so

   \[E_n = -\frac{13.6(2^2)}{n^2} = -\frac{54.4}{n^2}\ \mathrm{eV}.\]

   At \(n=3\),

   \[E_3 = -\frac{54.4}{9} = -6.04\ \mathrm{eV}.\]

   For a bound Coulomb orbit,

   \[K = -E, \qquad U = 2E.\]

   Therefore,

   \[K = 6.04\ \mathrm{eV},\]

   and

   \[U = 2(-6.04) = -12.09\ \mathrm{eV}.\]

   To ionize the electron from this state, we must raise its total energy from \(-6.04\ \mathrm{eV}\) to \(0\).

   So the minimum ionizing photon energy is

   \[E_{\mathrm{ion}} = 6.04\ \mathrm{eV}.\]

   
   Answer: (a) \(-6.04\ \mathrm{eV}\); (b) \(6.04\ \mathrm{eV}\) kinetic; (c) \(-12.09\ \mathrm{eV}\) potential; (d) \(6.04\ \mathrm{eV}\) to ionize.

4. ★★★
   Consider the Paschen series of the hydrogen-like ion \(\mathrm{He^+}\), so all transitions end at \(n_f=3\).

   • (a) Find the shortest wavelength in this series.
   • (b) Find the longest wavelength in this series.
   • (c) Find the wavelength of the specific transition \(n=5 \to n=3\).
   • (d) If the electron is in the \(n=3\) state of \(\mathrm{He^+}\), find its total energy, kinetic energy, and electric potential energy.
   Solution

   For a hydrogen-like ion,

   \[\frac{1}{\lambda} = R_H Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right).\]

   For \(\mathrm{He^+}\), \(Z=2\), so

   \[\frac{1}{\lambda} = 4R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right).\]

   Since this is the Paschen series, \(n_f=3\).

   The shortest wavelength comes from the largest energy transition in the series, which occurs for \(n_i\to\infty\):

   \[\frac{1}{\lambda_{\min}} = 4R_H\left(\frac{1}{9}-0\right)=\frac{4R_H}{9}.\]

   So

   \[\lambda_{\min}=\frac{9}{4R_H} = \frac{9}{4(1.097\times10^7)} = 2.05\times10^{-7}\ \mathrm{m}.\]

   Thus,

   \[\lambda_{\min} \approx 205\ \mathrm{nm}.\]

   The longest wavelength comes from the smallest energy transition in the series, namely \(n_i=4 \to n_f=3\):

   \[\frac{1}{\lambda_{\max}} = 4R_H\left(\frac{1}{9}-\frac{1}{16}\right) =4R_H\left(\frac{7}{144}\right)=\frac{7R_H}{36}.\]

   Therefore,

   \[\lambda_{\max}=\frac{36}{7R_H} = \frac{36}{7(1.097\times10^7)} = 4.69\times10^{-7}\ \mathrm{m}.\]

   So

   \[\lambda_{\max} \approx 469\ \mathrm{nm}.\]

   For the \(5\to3\) transition,

   \[\frac{1}{\lambda} = 4R_H\left(\frac{1}{9}-\frac{1}{25}\right) =4R_H\left(\frac{16}{225}\right)=\frac{64R_H}{225}.\]

   Hence,

   \[\lambda = \frac{225}{64R_H} = \frac{225}{64(1.097\times10^7)} = 3.20\times10^{-7}\ \mathrm{m}.\]

   Therefore,

   \[\lambda_{5\to3} \approx 320\ \mathrm{nm}.\]

   For the \(n=3\) state of \(\mathrm{He^+}\),

   \[E_3=-\frac{13.6Z^2}{n^2}=-\frac{13.6(4)}{9}=-6.04\ \mathrm{eV}.\]

   For a bound Coulomb orbit,

   \[K=-E, \qquad U=2E.\]

   So

   \[K=6.04\ \mathrm{eV}, \qquad U=-12.09\ \mathrm{eV}.\]

   
   Answer: (a) \(\lambda_{\min} \approx 205\ \mathrm{nm}\); (b) \(\lambda_{\max} \approx 469\ \mathrm{nm}\); (c) \(\lambda_{5\to3} \approx 320\ \mathrm{nm}\); (d) \(E=-6.04\ \mathrm{eV}\), \(K=6.04\ \mathrm{eV}\), \(U=-12.09\ \mathrm{eV}\).

[2.3] X-Rays and Nuclear Binding Energy⚓︎

1. ★☆☆
   An x-ray tube is operated at a potential difference of \(35.0\ \mathrm{kV}\).

   • (a) What is the maximum possible photon energy in the spectrum?
   • (b) What is the cutoff wavelength \(\lambda_0\)?
   Solution

   The largest possible photon energy occurs when one electron gives all of its kinetic energy to a single x-ray photon.

   Thus,

   \[E_{\max} = eV = 35.0\ \mathrm{keV}.\]

   In joules,

   \[E_{\max} = 3.50 \times 10^4\ \mathrm{eV}(1.602 \times 10^{-19}\ \mathrm{J/eV}) = 5.61 \times 10^{-15}\ \mathrm{J}.\]

   The cutoff wavelength is

   \[\lambda_0 = \frac{hc}{E_{\max}} = \frac{1240\ \mathrm{eV\cdot nm}}{3.50\times10^4\ \mathrm{eV}} = 0.0354\ \mathrm{nm}.\]

   
   Answer: (a) \(35.0\ \mathrm{keV}\), or \(5.61 \times 10^{-15}\ \mathrm{J}\); (b) \(\lambda_0 = 0.0354\ \mathrm{nm}\).

2. ★★☆
   In an x-ray tube, the accelerating voltage is held fixed, but the target metal is changed from molybdenum to tungsten.

   • (a) What happens to the cutoff wavelength \(\lambda_0\)?
   • (b) What happens to the characteristic peaks?
   • (c) What physical quantity controls each of these features?
   Solution

   The cutoff wavelength depends on the maximum kinetic energy of the electrons striking the target, and that is set by the accelerating voltage:

   \[E_{\max}=eV, \qquad \lambda_0 = \frac{hc}{eV}.\]

   Since the voltage is unchanged, the cutoff wavelength does not change.

   The characteristic peaks come from electrons making transitions between inner atomic energy levels of the target atoms. Those energy differences depend on the target material, so when the target changes from molybdenum to tungsten, the positions of the characteristic peaks change.

   
   Answer: (a) \(\lambda_0\) stays the same; (b) the characteristic peaks shift because the atomic energy-level spacings of the target change; (c) \(\lambda_0\) is controlled by the tube voltage, while the characteristic lines are controlled by the target material.

3. ★★★
   The neutral iron-56 atom \(\left(^{56}_{26}\mathrm{Fe}\right)\) has atomic mass \(55.934936\ \mathrm{u}\).

   • (a) How many protons and neutrons does it contain?
   • (b) Find the mass defect \(\Delta m\).
   • (c) Find the total binding energy of the nucleus.
   • (d) Find the average binding energy per nucleon.
   Solution

   The nucleus contains

   \[Z=26 \text{ protons}, \qquad N = 56-26 = 30 \text{ neutrons}.\]

   Using atomic masses, the mass defect is conveniently found from

   \[\Delta m = Zm_H + Nm_n - m_{\mathrm{atom}}.\]

   So

   \[\Delta m = 26(1.007825) + 30(1.008665) - 55.934936.\]

   This gives

   \[\Delta m = 0.528464\ \mathrm{u}.\]

   The total binding energy is

   \[BE = \Delta m(931.5\ \mathrm{MeV}/c^2)\]

   \[BE = (0.528464)(931.5)=492\ \mathrm{MeV}.\]

   The average binding energy per nucleon is

   \[\frac{BE}{A} = \frac{492}{56} = 8.79\ \mathrm{MeV/nucleon}.\]

   
   Answer: (a) 26 protons and 30 neutrons; (b) \(\Delta m = 0.528\ \mathrm{u}\); (c) \(BE \approx 492\ \mathrm{MeV}\); (d) \(8.79\ \mathrm{MeV/nucleon}\).

4. ★★★
   An x-ray tube is operated at \(50.0\ \mathrm{kV}\) using a tungsten target.

   • (a) Find the cutoff wavelength \(\lambda_0\).
   • (b) The voltage is then reduced until the cutoff wavelength shifts 25% to the right. What is the new voltage?
   • (c) At this new voltage, the target is changed from tungsten to molybdenum. What happens to the cutoff wavelength, and what happens to the characteristic peaks?
   Solution

   The cutoff wavelength is set by the maximum photon energy:

   \[E_{\max}=eV, \qquad \lambda_0 = \frac{hc}{eV}.\]

   For \(V=50.0\ \mathrm{kV}\),

   \[E_{\max}=50.0\ \mathrm{keV} = 5.00\times10^4\ \mathrm{eV}.\]

   Thus,

   \[\lambda_0 = \frac{1240\ \mathrm{eV\cdot nm}}{5.00\times10^4\ \mathrm{eV}} = 0.0248\ \mathrm{nm}.\]

   If the cutoff wavelength shifts 25% to the right, then

   \[\lambda_0' = 1.25\lambda_0.\]

   Since

   \[\lambda_0 \propto \frac{1}{V},\]

   the voltage must decrease by the same factor:

   \[V' = \frac{V}{1.25} = \frac{50.0\ \mathrm{kV}}{1.25} = 40.0\ \mathrm{kV}.\]

   At this new voltage, the cutoff wavelength is

   \[\lambda_0' = 1.25(0.0248\ \mathrm{nm}) = 0.0310\ \mathrm{nm}.\]

   Now change the target from tungsten to molybdenum while keeping the voltage fixed at \(40.0\ \mathrm{kV}\).

   The cutoff wavelength does not change, because it depends only on the accelerating voltage:

   \[\lambda_0 = \frac{hc}{eV}.\]

   However, the characteristic peaks do change, because they come from inner-shell transitions in the target atoms, and those energy differences depend on the target material.

   
   Answer: (a) \(\lambda_0 = 0.0248\ \mathrm{nm}\); (b) the new voltage is \(40.0\ \mathrm{kV}\); (c) the cutoff wavelength stays at \(0.0310\ \mathrm{nm}\) at the new voltage, but the characteristic peaks shift because the target material changed.