Answer
See explanation.
Work Step by Step
a. The largest energy photon has the shortest wavelength.
$E=\frac{hc}{\lambda}=\frac{(4.136\times10^{-15}eV)(3.00\times10^8 m/s)}{380\times10^{-9}m}=3.27eV$
The smallest energy photon has the longest wavelength.
$E=\frac{hc}{\lambda}=\frac{(4.136\times10^{-15}eV)(3.00\times10^8 m/s)}{750\times10^{-9}m}=1.65eV$
b. Only transitions with energy between 1.65 eV and 3.27 eV will emit/absorb visible-light photons. Figure 39.27 shows the energy levels for He+. The 6 to 4 (1.9 eV) and the 4 to 3 (2.6 eV) transitions are the only ones to qualify.
