Q. 50

# The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Given:

Wavelength, λ1 = 589 nm

Wavelength, λ2 = 589.6 nm

To find frequency:

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×108 m/s

Frequency, v1 = [3×108] / [589 × 10-9]

v1 = 5.093×1014 Hz

Frequency, v2 = [3×108] / [589.6 × 10-9]

V2 = 5.088×1014 Hz

Change in Energy, ∆E = E2-E1 = h[v2-v1]

= [6.626× 10-34] {[5.093 - 5.088] × 1014}

= [6.626× 10-34] {[5× 10-3] × 1014}

= 3.313× 10-22J

Therefore, the energy difference between two excited states is 3.31× 10-22J.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Structure of Atom - 0757 mins
Structure of Atom - 0666 mins
Structure of Atom - 0461 mins
Structure of Atom - 0366 mins
Structure of Atom - 0558 mins
Structure of Atom - 0868 mins
Understand photoelectric effect42 mins
Structure of Atom - 0165 mins
Structure of Atom - 0262 mins
Planck's Quantum theory65 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses