a. To determine whether the atom gains or loses energy for each transition, we can use the equation:
ΔE = E_final - E_initial
Where ΔE represents the change in energy, E_final is the energy of the final state, and E_initial is the energy of the initial state.
For each transition:
- ni = 6; nf = 4 : The atom loses energy because the final state has a lower energy level than the initial state.
- ni = 4; nf = 2 : The atom loses energy.
- ni = 5; nf = 3 : The atom loses energy.
- ni = 2; nf = 5 : The atom gains energy because the final state has a higher energy level than the initial state.
- ni = 5; nf = 7 : The atom gains energy.
b. To find the transition where the atom gains the most energy, we need to compare the energies of the initial and final states for each transition. Specifically, we want to find the transition with the largest difference in energy.
From the given transitions, the transition with the largest energy difference (gains the most energy) is ni = 2; nf = 5. To calculate the exact energy gained, we need to know the energy levels corresponding to ni and nf. In this case, the initial energy level (ni = 2) corresponds to the Lyman series, and the final energy level (nf = 5) corresponds to the Paschen series.
The equation to calculate the energy of a transition in the hydrogen atom is given by:
E = -13.6 eV * (Z^2 / n^2)
Where Z is the atomic number (1 for hydrogen) and n is the principal quantum number.
For ni = 2 (Lyman series):
E_initial = -13.6 eV * (1^2 / 2^2) = -13.6 eV * 0.25 = -3.4 eV
For nf = 5 (Paschen series):
E_final = -13.6 eV * (1^2 / 5^2) = -13.6 eV * 0.04 = -0.544 eV
Therefore, the atom gains energy:
ΔE = E_final - E_initial = (-0.544 eV) - (-3.4 eV) = 2.856 eV
c. To find the transition that will emit the shortest wavelength photon, we need to determine the corresponding energy levels and calculate the energy difference.
The energy of a photon is given by the equation:
E = hc / λ
Where E is the energy of the photon, h is Planck's constant (6.62607015 × 10^-34 J·s), c is the speed of light (2.998 × 10^8 m/s), and λ is the wavelength of the photon.
Since we want the shortest wavelength, we are looking for the transition with the highest energy difference.
From the given transitions, the transition that corresponds to the largest energy difference is ni = 6; nf = 4. To calculate the energy difference, we can use the same equation as in part b:
E_initial = -13.6 eV * (1^2 / 6^2) = -13.6 eV * 0.0278 = -0.377 eV
E_final = -13.6 eV * (1^2 / 4^2) = -13.6 eV * 0.0625 = -0.85 eV
ΔE = E_final - E_initial = (-0.85 eV) - (-0.377 eV) = -0.473 eV
Finally, we can calculate the corresponding wavelength using the energy of the photon:
E = -0.473 eV = (-0.473 eV) * (1.602 × 10^-19 J/eV) = -7.57 × 10^-20 J
λ = hc / E = (6.62607015 × 10^-34 J·s * 2.998 × 10^8 m/s) / (-7.57 × 10^-20 J)
λ ≈ 82.77 nm
Therefore, the transition from ni = 6 to nf = 4 will emit a photon with a wavelength of approximately 82.77 nm.