## University Physics with Modern Physics (14th Edition)

103 nm. $2.92\times10^{15}Hz$.
For a hydrogen atom, $E_n=-\frac{13.6eV}{n^2}$. Calculate the energy change $\Delta E = E_3-E_1=12.09eV$. The magnitude of the energy change for the atom can be related to the wavelength of the absorbed photon. $$\Delta E=hf=h\frac{c}{\lambda}$$ $$\lambda =\frac{hc}{\Delta E}=\frac{1.24\times10^{-6}eV\cdot m}{12.09eV}=1.03\times10^{-7}m$$ The frequency is $f=\frac{c}{\lambda}=2.92\times10^{15}Hz$.