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

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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.
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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