To determine the energy of one photon of visible radiation with a wavelength of [tex]\( 464.1 \, \text{nm} \)[/tex], we can use the formula for the energy of a photon, which is given by:
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
where:
- [tex]\( E \)[/tex] is the energy of the photon.
- [tex]\( h \)[/tex] is Planck’s constant, which is approximately [tex]\( 6.626 \times 10^{-34} \, \text{Js} \)[/tex].
- [tex]\( c \)[/tex] is the speed of light in a vacuum, which is [tex]\( 3.00 \times 10^8 \, \text{m/s} \)[/tex].
- [tex]\( \lambda \)[/tex] is the wavelength of the photon.
1. Convert the wavelength from nanometers to meters:
[tex]\[
\lambda = 464.1 \, \text{nm} = 464.1 \times 10^{-9} \, \text{m}
\][/tex]
2. Substitute the given values into the energy equation:
[tex]\[
E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3.00 \times 10^8 \, \text{m/s})}{464.1 \times 10^{-9} \, \text{m}}
\][/tex]
3. Calculate the numerator:
[tex]\[
6.626 \times 10^{-34} \, \text{Js} \times 3.00 \times 10^8 \, \text{m/s} = 1.9878 \times 10^{-25} \, \text{Jm}
\][/tex]
4. Divide by the wavelength in meters:
[tex]\[
\frac{1.9878 \times 10^{-25} \, \text{Jm}}{464.1 \times 10^{-9} \, \text{m}} = 4.28313 \times 10^{-19} \, \text{J}
\][/tex]
Therefore, the energy of one photon of visible radiation with a wavelength of [tex]\( 464.1 \, \text{nm} \)[/tex] is approximately [tex]\( 4.280 \times 10^{-19} \)[/tex] Joules.
Hence, the correct answer is:
[tex]\[
\boxed{4.280 \times 10^{-19} \, \text{J}}
\][/tex]
