Answer
\(4.78 \times 10^5 \mathrm{~J}\)
Work Step by Step
The energy of a single UV photon can be calculated using the equation:
\[ E = \frac{{hc}}{{\lambda}} \] where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\(6.62607015 \times 10^{-34} \mathrm{~J \cdot s}\)),
- \( c \) is the speed of light (\(3.00 \times 10^8 \mathrm{~m/s}\)),
- \( \lambda \) is the wavelength of the UV light in meters.
Given that the wavelength of the UV light is \(25 \mathrm{~nm}\), we need to convert it to meters by multiplying by \(10^{-9}\):
\[ \lambda = 25 \times 10^{-9} \mathrm{~m} = 2.5 \times 10^{-8} \mathrm{~m} \]
Now, we can calculate the energy of a single UV photon:
\[ E = \frac{{(6.62607015 \times 10^{-34} \mathrm{~J \cdot s}) \times (3.00 \times 10^8 \mathrm{~m/s})}}{{2.5 \times 10^{-8} \mathrm{~m}}} \]
\[ E \approx 7.95 \times 10^{-19} \mathrm{~J} \]
For 1 mole of UV photons (\(6.022 \times 10^{23}\) photons), the total energy can be calculated by multiplying the energy of a single photon by Avogadro's number:
\[ \text{Total energy} = (7.95 \times 10^{-19} \mathrm{~J/photon}) \times (6.022 \times 10^{23} \mathrm{~photons}) \]
\[ \text{Total energy} \approx 4.78 \times 10^5 \mathrm{~J} \]
So, the energy of 1 mole of UV photons with a wavelength of \(25 \mathrm{~nm}\) is approximately \(4.78 \times 10^5 \mathrm{~J}\).
