Problem • What is the frequency of light emitted from an electron in the hydrogen atom going from the 3rd energy level to the first. E = hv 1 AE = -RH(n? n? H = 6.63 x 10-34 Js Ry = 2.18 x 10-18 J
Problem • What is the frequency of light emitted from an electron in the hydrogen atom going from the 3rd energy level to the first. E = hv 1 AE = -RH(n? n? H = 6.63 x 10-34 Js Ry = 2.18 x 10-18 J
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![**Problem:**
What is the frequency of light emitted from an electron in the hydrogen atom going from the 3rd energy level to the first.
**Formulas:**
\[ \Delta E = -R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
\[ E = h\nu \]
**Constants:**
H = \( 6.63 \times 10^{-34} \) Js
\[ R_H = 2.18 \times 10^{-18} \) J
---
**Explanation of Formulas:**
1. **Energy Difference (\( \Delta E \))**:
The first formula calculates the change in energy (\( \Delta E \)) when an electron transitions between two energy levels in a hydrogen atom. Here:
- \( R_H \) is the Rydberg constant for hydrogen.
- \( n_f \) (final) and \( n_i \) (initial) are the principal quantum numbers of the final and initial states, respectively.
2. **Planck's Equation (\( E = h\nu \))**:
The second formula relates the energy (\( E \)) of a photon to its frequency (\( \nu \)):
- \( h \) is Planck's constant.
- \( \nu \) is the frequency of the emitted light.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6746fa6f-1856-443b-addc-ecae5d9da9b9%2F59e63952-b4d7-4eb1-8815-a57557800556%2Fxgrwojg.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
What is the frequency of light emitted from an electron in the hydrogen atom going from the 3rd energy level to the first.
**Formulas:**
\[ \Delta E = -R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
\[ E = h\nu \]
**Constants:**
H = \( 6.63 \times 10^{-34} \) Js
\[ R_H = 2.18 \times 10^{-18} \) J
---
**Explanation of Formulas:**
1. **Energy Difference (\( \Delta E \))**:
The first formula calculates the change in energy (\( \Delta E \)) when an electron transitions between two energy levels in a hydrogen atom. Here:
- \( R_H \) is the Rydberg constant for hydrogen.
- \( n_f \) (final) and \( n_i \) (initial) are the principal quantum numbers of the final and initial states, respectively.
2. **Planck's Equation (\( E = h\nu \))**:
The second formula relates the energy (\( E \)) of a photon to its frequency (\( \nu \)):
- \( h \) is Planck's constant.
- \( \nu \) is the frequency of the emitted light.
Transcribed Image Text:## Examples
1. **Calculate the frequency of light with a wavelength of 585 nm.**
2. **Calculate the wavelength of light with a frequency of 1.89 x 10¹⁸ Hz. A Hz is s⁻¹**
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