## College Physics (4th Edition)

$\lambda = 368~nm$
We can find the energy of the ultraviolet photon: $E = \frac{hc}{\lambda}$ $E = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{320\times 10^{-9}~m}$ $E = (6.212\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $E = 3.88~eV$ We can find the energy $E_p$ of the emitted photon: $E_p = 3.88~eV-0.500~eV = 3.38~eV$ We can find the wavelength of the emitted photon: $E_p = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{E_p}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(3.38~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda = 3.68\times 10^{-7}~m$ $\lambda = 368~nm$