An electron accelerates through a 12.5V potential difference, starting from rest, and then collides with a hydrogen atom, exciting the atom to the highest energy level allowed. List all the possible quantum-jump transitions by which the excited atom could emit a photon and the wavelength (in nm) of each.

Short Answer

Expert verified

The wavelength isλ1=656.08nm,λ2=102.56nmandλ3=121.56nm.

Step by step solution

01

Given Information

We need to find the emit a photon and the wavelength.

02

Simplify

The atom could emit a photon in the 3possible quantum-jump transition:

role="math" localid="1650276055698" 32,31and21

Now, for 32:

E1=-1.51eV--3.40eVE1=-1.51eV+3.40eVE1=1.89eV

for 31:

E2=-1.51eV--13.60eVE2=-1.51eV+13.60eVE2=12.09eV

for 21:

E3=-3.40eV--13.60eVE3=-3.40eV+13.60eVE3=10.2eV

03

Calculation

Now, finding the wavelength :

λ1=hcE1λ1=12.40eV×nm1.89eVλ1=656.08nmλ2=hcE2λ2=12.40eV×nm12.09eVλ2=102.56nmλ3=hcE3λ3=12.40eV×nm10.2eVλ3=121.56nm

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

Study anywhere. Anytime. Across all devices.

Sign-up for free