asagaradze40
Answered

Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

3) A photon is emitted as the electron in a hydrogen
atom drops from the n = 5 energy level directly to
the n = 3 energy level. What is the energy of the
emitted photon?
(1) 0.85 eV
(2) 0.97 eV
(3) 1.51 eV
(4) 2.05 eV

Sagot :

pstnonsonjoku

From the information provided and the calculations, the energy of the drop is 1.51 eV

What is the Rydberg formula?

The Rydberg formula is used to obtain the energy of a photon or the wavelength of a photon.

Thus;

1/λ = 1.097 * 10^7 (1/3^2 - 1/5^2)

1/λ = 1.097 * 10^7(0.11 - 0.04)

λ = 1.3 * 10^-6 m

Since;

E = hc/λ

E = 6.6 * 10^-34 * 3 * 10^8/1.3 * 10^-6

E = 1.51 * 10^-19 J or 1.51 eV

Learn more about photon:https://brainly.com/question/20912241

#SPJ1

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.