Let's solve the problem step by step.
### Given Data:
- Wavelength of sodium absorption line, [tex]\(\lambda = 589 \, \text{nm}\)[/tex]
- Speed of light in a vacuum, [tex]\(c = 3.00 \times 10^8 \, \text{m/s}\)[/tex]
- Planck's constant, [tex]\(h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)[/tex]
### Step-by-Step Solution:
1. Convert the wavelength from nanometers to meters:
- [tex]\(1 \, \text{nm} = 10^{-9} \, \text{m}\)[/tex]
- [tex]\(\lambda = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m} = 5.89 \times 10^{-7} \, \text{m}\)[/tex]
2. Calculate the frequency ([tex]\( \nu \)[/tex]) of the light using the formula:
[tex]\[
\nu = \frac{c}{\lambda}
\][/tex]
- Plug in the values:
[tex]\[
\nu = \frac{3.00 \times 10^8 \, \text{m/s}}{5.89 \times 10^{-7} \, \text{m}}
\][/tex]
- Solving for [tex]\(\nu\)[/tex]:
[tex]\[
\nu = 5.0933786078098475 \times 10^{14} \, \text{Hz}
\][/tex]
3. Calculate the energy (E) of the photon using Planck's equation:
[tex]\[
E = h \nu
\][/tex]
- Plug in the values:
[tex]\[
E = (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (5.0933786078098475 \times 10^{14} \, \text{Hz})
\][/tex]
- Solving for [tex]\(E\)[/tex]:
[tex]\[
E = 3.374872665534805 \times 10^{-19} \, \text{J}
\][/tex]
### Conclusion:
The energy of the sodium absorption line at [tex]\(589 \, \text{nm}\)[/tex] is approximately:
[tex]\[
3.37 \times 10^{-19} \, \text{J}
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{3.37 \times 10^{-19} \, \text{J}}
\][/tex]
Which corresponds to option C.
