An electron absorbs a single photon with a wavelength = 127 nm. Calculate the energy of the photon. O 1.56 x 1032 J O 1.56 x 10-18 J O 1.56 x 10-27 J O 6.39 × 10-26 j
An electron absorbs a single photon with a wavelength = 127 nm. Calculate the energy of the photon. O 1.56 x 1032 J O 1.56 x 10-18 J O 1.56 x 10-27 J O 6.39 × 10-26 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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An electron absorbs a single photon with a wavelength = 127 nm. Calculate the energy of the photon.
![**Question:** An electron absorbs a single photon with a wavelength = 127 nm. Calculate the energy of the photon.
**Options:**
- ○ 1.56 × 10³² J
- ○ 1.56 × 10⁻¹⁸ J
- ○ 1.56 × 10⁻²⁷ J
- ○ 6.39 × 10⁻²⁶ J
**Explanation:**
To calculate the energy of a photon, you can use the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( E \) is the energy,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\)),
- \( c \) is the speed of light (\(3.00 \times 10^8 \, \text{m/s}\)),
- \( \lambda \) is the wavelength (\(127 \, \text{nm} = 127 \times 10^{-9} \, \text{m}\)).
Substituting the known values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J}\cdot\text{s})(3.00 \times 10^8 \, \text{m/s})}{127 \times 10^{-9} \, \text{m}} \]
By calculating, you should match one of the given options.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F840f535b-f075-49f5-9fe8-c5a30f6f8dd7%2Fcf787254-abe5-4f76-8c90-ae190361b9ff%2Fb7nsc2g_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:** An electron absorbs a single photon with a wavelength = 127 nm. Calculate the energy of the photon.
**Options:**
- ○ 1.56 × 10³² J
- ○ 1.56 × 10⁻¹⁸ J
- ○ 1.56 × 10⁻²⁷ J
- ○ 6.39 × 10⁻²⁶ J
**Explanation:**
To calculate the energy of a photon, you can use the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( E \) is the energy,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\)),
- \( c \) is the speed of light (\(3.00 \times 10^8 \, \text{m/s}\)),
- \( \lambda \) is the wavelength (\(127 \, \text{nm} = 127 \times 10^{-9} \, \text{m}\)).
Substituting the known values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J}\cdot\text{s})(3.00 \times 10^8 \, \text{m/s})}{127 \times 10^{-9} \, \text{m}} \]
By calculating, you should match one of the given options.
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