Answer
$\lambda = 2.1\times 10^{-6}~m$
Work Step by Step
We can express $0.6~eV$ in units of joules:
$(0.6~eV)(\frac{1.6\times 10^{-19}~J}{1~eV}) = 9.6\times 10^{-20}~J$
We can find the greatest wavelength of light that can be recorded:
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.63\times 10^{-34}~J\cdot s)(3.0\times 10^8~m/s)}{9.6\times 10^{-20}~J}$
$\lambda = 2.1\times 10^{-6}~m$
