Atoms — Questions & Answers

Sample questions from Atoms

1. Q1. An electron of kinetic energy 12.09 eV has a de Broglie wavelength of about 0.353 nm, while the photon that carries the same energy has a wavelength of about 102.6 nm. The reason the two differ so greatly is:

   • A.the electron is charged and the photon is not
   • B.the de Broglie relation does not apply to photons
   • C.for a massive particle p = sqrt(2mK), whereas for a photon p = E/c; these give quite different momenta at the same energy✓
   • D.the electron wavelength was measured in a different medium
   Solution

   Both obey lambda = h/p, but the energy-momentum relations differ. A 12.09 eV electron has p = sqrt(2 x 9.1x10^-31 x 12.09 x 1.6x10^-19) = 1.88x10^-24 kg m/s, while the photon has p = E/c = 6.45x10^-27 kg m/s, some 290 times smaller, hence a much longer wavelength.

2. Q2. A hydrogen atom's electron transitions from a HIGH excited state (say n=10) down through MANY successive intermediate single-step transitions (n=10 to n=9, then n=9 to n=8, and so on, all the way down to n=1), rather than making one single direct n=10-to-n=1 jump. Comparing the TOTAL energy released across all these many small successive steps to the energy that WOULD be released by one single direct n=10-to-n=1 transition, the total energy released is:

   • A.EXACTLY THE SAME in both cases (conservation of energy guarantees the total energy released, summed across any sequence of intermediate steps between the SAME starting and ending levels, must equal the energy released by one single direct transition between those same two levels), though the many small steps produce MULTIPLE lower-energy photons, while the single direct transition produces ONE higher-energy photon✓
   • B.LESS energy released via the many small steps, compared to the single direct transition
   • C.MORE energy released via the many small steps, compared to the single direct transition
   • D.The comparison is undefined, since these represent fundamentally different, incomparable physical processes
   Solution

   By conservation of energy (a fundamental physical principle that must hold regardless of the SPECIFIC pathway/sequence of intermediate steps taken), the TOTAL energy released in transitioning from n=10 down to n=1 must be EXACTLY THE SAME (equal to E10-E1, the total energy difference between these two specific starting and ending levels), REGARDLESS of whether this transition occurs via one single direct step, or via many smaller successive intermediate steps -- what DOES differ between these two scenarios is HOW this total energy is distributed: the single direct n=10-to-n=1 transition releases this entire total energy as ONE single, relatively high-energy photon, while the many-small-steps pathway instead releases this SAME total energy distributed across MULTIPLE separate, individually LOWER-energy photons (one photon for each individual intermediate step, e.g. n=10-to-n=9, n=9-to-n=8, and so on) -- summing the individual energies of all these multiple lower-energy photons gives EXACTLY the same total energy as the single high-energy photon from the direct transition -- this illustrates an important, general principle: while the SPECIFIC decay PATHWAY (which particular intermediate levels are visited, and in what order/how many steps) can genuinely vary between different individual atoms (or even for the same atom on different occasions), the TOTAL energy eventually released, between any two SPECIFIC starting and ending energy levels, remains invariant, fixed by simple energy conservation, regardless of the specific pathway taken. A student who assumes the total energy released differs between these two pathway scenarios picks B or C, incorrectly, violating energy conservation.

3. Q3. Hydrogen atoms in the ground state are bombarded by a monoenergetic electron beam. The beam energy is slowly raised. The first sign of emitted light appears when the beam energy reaches:

   • A.1.89 eV
   • B.13.6 eV
   • C.10.2 eV✓
   • D.3.4 eV
   Solution

   An electron can transfer any amount of energy up to its full kinetic energy, but the atom can accept only a level difference. The smallest one from n = 1 is E2 - E1 = 10.2 eV, and the resulting 2-to-1 decay gives the 121.6 nm Lyman-alpha photon. This threshold behaviour is the essence of the Franck-Hertz experiment.

4. Q4. Hydrogen atoms are excited by photons of a single energy and the emission spectrum then shows exactly 10 different lines. The photon energy was:

   • A.13.06 eV✓
   • B.12.75 eV
   • C.12.09 eV
   • D.13.6 eV
   Solution

   10 lines requires n(n-1)/2 = 10, so n = 5. Raising ground-state atoms to n = 5 needs E5 - E1 = 13.6(1 - 1/25) = 13.056 eV. 12.75 eV reaches n = 4 (6 lines) and 12.09 eV reaches n = 3 (3 lines); 13.6 eV would ionize.

5. Q5. The wave number (1/λ) of the H-beta line (n = 4 → 2) of hydrogen is approximately (R = 1.097×10^7 m^-1):

   • A.8.23×10^6 m^-1
   • B.2.06×10^6 m^-1✓
   • C.1.52×10^6 m^-1
   • D.2.74×10^6 m^-1
   Solution

   1/λ = R(1/2^2 - 1/4^2) = 1.097×10^7 × (1/4 - 1/16) = 1.097×10^7 × 3/16 = 2.06×10^6 m^-1, giving λ = 486 nm.

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