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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2. Calculate the energy associated with the light emitted by Rubidium and Cesium
![### Ionizing Flame Colors and Energy Estimation
When certain elements are heated, they emit specific colors due to the energy released as electrons transition between more stable energy levels. The color produced when an electron drops to a lower energy state within the visible portion of the electromagnetic spectrum is unique to each element. This phenomenon is often used to identify elements in various substances.
#### Estimating Energy of Light Emission
The energy emitted by an element can be calculated using Planck’s equation for energy, \(E = \frac{hv}{\lambda}\), where:
- \(h\) is Planck’s constant \((6.63 \times 10^{-34} \text{J} \cdot \text{s})\),
- \(c\) is the speed of light in a vacuum \((2.998 \times 10^8 \text{m/s})\),
- \(v\) is the frequency in \(\text{s}^{-1}\),
- \(\lambda\) is the wavelength in meters.
**Example Calculation for Arsenic:**
1. **Convert Wavelength to Meters (Equation 1):**
\[
\lambda = 419 \, \text{nm} \times \frac{1 \, \text{m}}{1 \times 10^9 \, \text{nm}} = 4.19 \times 10^{-7} \, \text{m}
\]
2. **Calculate Energy (Equation 2):**
\[
E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34} \, \text{J} \cdot \text{s}) (2.998 \times 10^8 \, \text{m/s})}{4.19 \times 10^{-7} \, \text{m}} = 4.19 \times 10^{-19} \, \text{J}
\]
#### Table 1: Cations and Colors in Flame
This table lists various cations, the colors they produce when ionized in a flame, and their corresponding wavelengths in nanometers.
| Symbol | Name | Color | Wavelength, nm |
|--------|----------------|---------------|----------------|
| As | Arsenic | Blue | 419 |
| B | Boron | Bright Green](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98cfde8d-018b-4525-8f81-df4d401b811f%2F49e2e7cc-5477-4fdc-bb49-ab5aaddcce0a%2Fxw7op67.jpeg&w=3840&q=75)
Transcribed Image Text:### Ionizing Flame Colors and Energy Estimation
When certain elements are heated, they emit specific colors due to the energy released as electrons transition between more stable energy levels. The color produced when an electron drops to a lower energy state within the visible portion of the electromagnetic spectrum is unique to each element. This phenomenon is often used to identify elements in various substances.
#### Estimating Energy of Light Emission
The energy emitted by an element can be calculated using Planck’s equation for energy, \(E = \frac{hv}{\lambda}\), where:
- \(h\) is Planck’s constant \((6.63 \times 10^{-34} \text{J} \cdot \text{s})\),
- \(c\) is the speed of light in a vacuum \((2.998 \times 10^8 \text{m/s})\),
- \(v\) is the frequency in \(\text{s}^{-1}\),
- \(\lambda\) is the wavelength in meters.
**Example Calculation for Arsenic:**
1. **Convert Wavelength to Meters (Equation 1):**
\[
\lambda = 419 \, \text{nm} \times \frac{1 \, \text{m}}{1 \times 10^9 \, \text{nm}} = 4.19 \times 10^{-7} \, \text{m}
\]
2. **Calculate Energy (Equation 2):**
\[
E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34} \, \text{J} \cdot \text{s}) (2.998 \times 10^8 \, \text{m/s})}{4.19 \times 10^{-7} \, \text{m}} = 4.19 \times 10^{-19} \, \text{J}
\]
#### Table 1: Cations and Colors in Flame
This table lists various cations, the colors they produce when ionized in a flame, and their corresponding wavelengths in nanometers.
| Symbol | Name | Color | Wavelength, nm |
|--------|----------------|---------------|----------------|
| As | Arsenic | Blue | 419 |
| B | Boron | Bright Green
Transcribed Image Text:**Transcription for Educational Website**
---
### Spectroscopy and Light Emission
1. **List the expected color of light and the corresponding wavelength (nm) associated with each cation (use Table 1).**
- **Rubidium, Rb⁺**
- Color: Pale violet
- Wavelength: 406 nm
- **Cesium, Cs⁺**
- Color: Pale violet
- Wavelength: 403 nm
2. **Calculate the energy associated with the light emitted by each of the elements above (use equations 1 and 2).**
- **Rubidium, Rb⁺**
- *(Calculation section left blank)*
- **Cesium, Cs⁺**
- *(Calculation section left blank)*
---
Note: The image does not contain any graphs or diagrams, only text and sections for calculations.
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