Final answer:
The atom that produces the shortest wavelength spectral line during the transition from n=2 to n=1 in the Lyman series is the double ionized lithium (Li++), as it has a higher effective atomic number (Z) after accounting for electron count, leading to a greater energy change and consequently a shorter wavelength photon.
Step-by-step explanation:
The question is asking which atom produces the shortest wavelength spectral line when an electron transitions from the n=2 energy level to the n=1 level. This is a reference to the Lyman series in the hydrogen atom spectrum, which deals with the emissions of photons by electrons in excited states transitioning to the ground state (n=1). The energy difference and thus the frequency (and wavelength) of the emitted photon depend on the atomic number (Z) of the hydrogen-like atom because the energies of the orbits in a one-electron atom scale with the square of Z. Therefore, the greater Z is, the more energy is released and the shorter the wavelength of the emitted photon.
For a hydrogen atom (Z=1), a deuterium atom (also Z=1, but with an extra neutron), a singly ionized lithium (Li+, Z=3), and a double ionized lithium (Li++, Z=3 but with one less electron than Li+), transitions from n=2 to n=1 will produce lines in the Lyman series. As Z increased, the energy of the photon increases, and its wavelength decreases due to the E=hc/λ relationship, where E is energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Considering the atoms in question, double ionized lithium, which has a higher Z value than hydrogen or deuterium after accounting for the number of electrons, will have the most significant decrease in energy levels, resulting in the emission of a photon with the shortest wavelength.