Answer: \( E \approx 3.29 \times 10^{-19} \, \text{J} \)
Explanation: The energy of a photon can be calculated using the formula:
\[
E = \frac{hc}{\lambda}
\]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck’s constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.
First, convert the wavelength from angstroms to meters:
\[
6057.8 \, \text{Å} = 6057.8 \times 10^{-10} \, \text{m}
\]
Now, substitute the values into the energy formula:
\[
E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{6057.8 \times 10^{-10} \, \text{m}}
\]
Calculating this gives:
\[
E \approx 3.29 \times 10^{-19} \, \text{J}
\]