The first vibrational level for NaH lies at 1.154 × 10^-20 J and the second vibrational level lies at 3.406 × 10^-20 J.What is the frequency of the photon emitted when a molecule of NaH drops from the second vibrational level to the first vibrational level? A)1)742 × 10¹³ Hz B)3)399 × 10¹³ Hz C)5)140 × 10¹³ Hz D)6)882 × 10¹³ Hz

Question

Quizplus · Accepted Answer

The energy difference between the second and first vibrational levels can be found as follows:

ΔE = E2 - E1 = (3.406 × 10^-20 J) - (1.154 × 10^-20 J) = 2.252 × 10^-20 J

The frequency of the photon emitted during this transition can be found using the equation:

ΔE = hν

where h is the Planck constant and ν is the frequency.

ν = ΔE/h = (2.252 × 10^-20 J) / (6.626 × 10^-34 J s) = 3.399 × 10^13 Hz

Therefore, the correct answer is B: 3.399 × 10^13 Hz.

The First Vibrational Level for NaH Lies at 1