AAMC MCAT Practice Exam 2

C/P: What expression gives the amount of light energy (in J per photon) that is converted to other forms between the fluorescence excitation and emission events? "intensity of fluorescence emission at 440 nm excitation at 360 nm) was monitored for 20 minutes" A) (6.62 × 10-34) × (3.0 × 108) B) (6.62 × 10-34) × (3.0 × 108) × (360 × 10-9) C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)] D) (6.62 × 10-34) × (3.0 × 108) / (440 × 10-9) - ANSWER C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)] The answer to this question is C because the equation of interest is E = hf = hc/λ, where h = 6.62 × 10 −34 J ∙ s and c = 3 × 10 8 m/s. Excitation occurs at λe = 360 nm, but fluorescence is observed at λf = 440 nm. This implies that an energy of E = (6.62 × 10 −34) × (3 × 10 8) × [1 / (360 × 10 −9) − 1 / (440 × 10 −9)] J per photon is converted to other forms between the excitation and fluorescence events