The electronic transition in Na from 3p1to3s1 gives rise to a bright yellow-orange emission at 589.2 nm. What is the energy of this transition?

Short Answer

Expert verified

The energy of this transition is3.371×10- 19J.

Step by step solution

01

Convert wavelengths in nanometres (nm) to metres (m).

The following equation is used to determine the energy of the transition:

E=hcλ______________1

E is the energy of transition, h is the Plank's constant, c is the light velocity and λis the emission wavelength.

Plank's constant has a value of6.26×10-34J.s.

The value of light velocity is2.99792×108m/s.

The emission wavelength is 589.2 nm.

Convert wavelengths in nanometres (nm) to metres (m).

λ=589.2nm×1m109nm=589.2×10-9m

02

Evaluation of energy of transition

Substitute all known values in equation (1).

E=6.26×10-34J.s×2.99792×108m/s589.2×10-9m=3.371×10-19J

Therefore, the energy of transition is 3.371×10-19J.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free