Answer
The cutoff wavelength is $~~296~nm$
Work Step by Step
It is given that $4.20~eV$ is required to eject electrons. We can express this energy in units of joules:
$E = (4.20~eV)(\frac{1.6\times 10^{-19}~J}{1~eV}) = 6.72\times 10^{-19}~J$
We can find the cutoff wavelength:
$E = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.63\times 10^{-34}~J\cdot s)(3.0\times 10^8~m/s)}{6.72\times 10^{-19}~J}$
$\lambda = 2.96\times 10^{-7}~m$
$\lambda = 296~nm$
The cutoff wavelength is $~~296~nm$
