QUESTION 21 Excited sodium atoms emit light in the infrared at 589 nm. What is the energy of a single photon with this wavelength? O 5.09 x 1014, O 1,12 x 10-27, O 3.37 x 10-19, O 3.37 x 10-28 O 1.30 x 10-19
QUESTION 21 Excited sodium atoms emit light in the infrared at 589 nm. What is the energy of a single photon with this wavelength? O 5.09 x 1014, O 1,12 x 10-27, O 3.37 x 10-19, O 3.37 x 10-28 O 1.30 x 10-19
Chemistry
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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
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![### Question 21
**Excited sodium atoms emit light in the infrared at 589 nm. What is the energy of a single photon with this wavelength?**
1. \( 5.09 \times 10^{14} \) J
2. \( 1.12 \times 10^{-27} \) J
3. \( 3.37 \times 10^{-19} \) J
4. \( 3.37 \times 10^{-28} \) J
5. \( 1.30 \times 10^{-19} \) J
To calculate the energy \( E \) of a single photon, use the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) J·s)
- \( c \) is the speed of light (\( 3.0 \times 10^8 \) m/s)
- \( \lambda \) is the wavelength (\( 589 \times 10^{-9} \) m)
By inserting the given wavelength into the formula, you can determine the energy of the photon corresponding to the given wavelength.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb8f64bb-a739-44c8-9c71-41ea58cbca19%2Fd3f523c0-18f3-47aa-973c-f1aa294563b1%2Fd0gqxca_processed.png&w=3840&q=75)
Transcribed Image Text:### Question 21
**Excited sodium atoms emit light in the infrared at 589 nm. What is the energy of a single photon with this wavelength?**
1. \( 5.09 \times 10^{14} \) J
2. \( 1.12 \times 10^{-27} \) J
3. \( 3.37 \times 10^{-19} \) J
4. \( 3.37 \times 10^{-28} \) J
5. \( 1.30 \times 10^{-19} \) J
To calculate the energy \( E \) of a single photon, use the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) J·s)
- \( c \) is the speed of light (\( 3.0 \times 10^8 \) m/s)
- \( \lambda \) is the wavelength (\( 589 \times 10^{-9} \) m)
By inserting the given wavelength into the formula, you can determine the energy of the photon corresponding to the given wavelength.
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