Answer
$\lambda = 368~nm$
Work Step by Step
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:
$E_p = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{E_p}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(3.38~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda = 3.68\times 10^{-7}~m$
$\lambda = 368~nm$
