Energy released in the hydrogen transition n = 3→1 falls on a photocathode. Later the same photocathode is exposed to the spectrum of an unknown hydrogen-like gas excited to n = 2. The de Broglie wavelength of the fastest emitted photoelectrons becomes one-third of its previous value. For this gas, the difference between the energies of the second Lyman line and the first Balmer line is three times the ionization potential of hydrogen. Select the correct statement(s).
Select all that apply

Question

Energy released in the hydrogen transition n = 3→1 falls on a photocathode. Later the same photocathode is exposed to the spectrum of an unknown hydrogen-like gas excited to n = 2. The de Broglie wavelength of the fastest emitted photoelectrons becomes one-third of its previous value. For this gas, the difference between the energies of the second Lyman line and the first Balmer line is three times the ionization potential of hydrogen. Select the correct statement(s).

Select all that apply

G1 Physics · Accepted Answer

For hydrogen, the 3→1 photon energy is 13.6(1−1/9) = 12.09 eV, so the first photoelectron kinetic energy is K₁ = 12.09−φ. The new de Broglie wavelength is one-third, so momentum is three times and K₂ = 9K₁. Using the spectral condition given for the unknown hydrogen-like gas identifies Z = 2 (He⁺). Solving the photoelectric-energy relation then gives φ = 8.5 eV. Correct statements: B and C.