Answer
$\lambda = 170~nm$
Work Step by Step
We can find the wavelength of the incident light:
$E = \frac{hc}{\lambda} = K_{max}+ \Phi$
$\frac{hc}{\lambda} = (1e)(5.0~V)+(2.2~eV)$
$\frac{hc}{\lambda} = 7.2~eV$
$\frac{hc}{\lambda} = (7.2~eV)(\frac{1.6\times 10^{-19}~J}{1~eV})$
$\frac{hc}{\lambda} = 1.152\times 10^{-18}~J$
$\lambda = \frac{hc}{1.152\times 10^{-18}~J}$
$\lambda = \frac{(6.63\times 10^{-34}~J\cdot s)(3.0\times 10^8~m/s)}{1.152\times 10^{-18}~J}$
$\lambda = 170~nm$
