A fluorescent solid absorbs a photon of ultraviolet light of wavelength 320 nm. If the solid dissipates 0.500 eV of the energy and emits the rest in a single photon, what is the wavelength of the emitted light?

To find the energy of the absorbed photon, we can use the Planck's relation, which states that the energy of a photon is given by \(E = h\cdot f\) where \(h\) is the Planck's constant (6.626 x 10^{-34} Js) and \(f\) is the frequency of the photon. We are not given the frequency, but we know the wavelength \(\lambda\), and we can use the relation between frequency and wavelength: \(f = \frac{c}{\lambda}\), where \(c\) is the speed of light (2.998 x 10^8 m/s). First, convert the wavelength from nm to meters: \(\lambda = 320\ nm = 320 \cdot 10^{-9}\ m\). Next, find the frequency: \(f = \frac{c}{\lambda} = \frac{2.998 \cdot 10^8\ m/s}{320 \cdot 10^{-9}\ m} = 9.36875 \cdot 10^{14}\ Hz\). Now, find the energy of the absorbed photon: \(E = h\cdot f = (6.626 \cdot 10^{-34}\ Js)(9.36875 \cdot 10^{14}\ Hz) = 6.2097 \cdot 10^{-19}\ J\).

We know that the solid dissipates 0.500 eV of energy. First, we need to convert this energy into Joules. 1 eV is equal to 1.602 x 10^{-19} J, so 0.500 eV is equal to \(0.500 \cdot (1.602 \cdot 10^{-19}\ J) = 8.01 \cdot 10^{-20}\ J\). Now, subtract the dissipated energy from the total absorbed energy to find the energy of the emitted photon: \(E_{emitted} = E_{absorbed} - E_{dissipated} = 6.2097 \cdot 10^{-19}\ J - 8.01 \cdot 10^{-20}\ J = 5.4087 \cdot 10^{-19}\ J\).

Now we have the energy of the emitted photon. We can use the Planck's relation again to find the frequency of the emitted photon: \(f_{emitted} = \frac{E_{emitted}}{h} = \frac{5.4087 \cdot 10^{-19}\ J}{6.626 \cdot 10^{-34}\ Js} = 8.15975 \cdot 10^{14}\ Hz\). Finally, use the relation between frequency and wavelength to find the wavelength of the emitted photon: \(\lambda_{emitted} = \frac{c}{f_{emitted}} = \frac{2.998 \cdot 10^8\ m/s}{8.15975 \cdot 10^{14}\ Hz} = 3.676 \cdot 10^{-7}\ m\). Convert the wavelength back to nm: \(\lambda_{emitted} = 3.676 \cdot 10^{-7}\ m = 367.6\ nm\). The wavelength of the emitted light is approximately \(367.6\ nm\).