When Cu^{2+} ions in water absorb visible light, the absorbance typically occurs around a wavelength of 800 nm. What is the energy of this absorbance in electron volts (eV)?

• A. 2.48 eV
• B. 1.55 eV
• C. 0.62 eV
• D. 3.10 eV

Answer: (B)

Verified step by step guidance

1. Identify the relationship between the energy of a photon and its wavelength. The energy \(E\) of a photon can be calculated using the equation \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the light.
2. Write down the known constants: Planck's constant \(h = 6.626 \times 10^{-34}\) J\cdot s, speed of light \(c = 3.00 \times 10^{8}\) m/s, and the given wavelength \(\lambda = 800\) nm. Remember to convert the wavelength from nanometers to meters by multiplying by \$10^{-9}$.
3. Substitute the values into the energy equation: \(E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{800 \times 10^{-9}}\) to find the energy in joules.
4. Convert the energy from joules to electron volts (eV) using the conversion factor \$1 \text{ eV} = 1.602 \times 10^{-19}\( joules. Use the formula \)E(\text{eV}) = \frac{E(\text{J})}{1.602 \times 10^{-19}}$.
5. Calculate the final value to find the energy of the absorbance in electron volts, which corresponds to the energy absorbed by the \(\mathrm{Cu^{2+}}\) ions at 800 nm.