## College Physics (4th Edition)

(a) $\lambda = 400~nm$ (b) $f = 7.5\times 10^{14}~Hz$
(a) We can convert the energy to units of joules: $E = 3.1~eV\times \frac{1.6\times 10^{-19}~J}{1~eV} = 4.96\times 10^{-19}~J$ We can find the wavelength: $E = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{E}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{4.96\times 10^{-19}~J}$ $\lambda = 4.0\times 10^{-7}~m$ $\lambda = 400~nm$ (b) We can find the frequency: $\lambda~f = c$ $f = \frac{c}{\lambda}$ $f = \frac{3.0\times 10^8~m/s}{4.0\times 10^{-7}~m}$ $f = 7.5\times 10^{14}~Hz$