Answer
The maximum wavelength of photons that will eject electrons from this metal is $507~nm$
Work Step by Step
We can find the magnitude of the energy removed from the electron by the stopping potential:
$E = q~\Delta V$
$E = (1.6\times 10^{-19}~C)(1.10~V)$
$E = 1.76\times 10^{-19}~J$
This energy is equal to the maximum kinetic energy of the electron as it is ejected from the metal plate. We can find the work function $\phi$:
$K_{max} = \frac{hc}{\lambda}-\phi$
$\phi = \frac{hc}{\lambda}-K_{max}$
$\phi = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~/s)}{350\times 10^{-9}~m}-1.76\times 10^{-19}~J$
$\phi = 5.679\times 10^{-19}~J-1.76\times 10^{-19}~J$
$\phi = 3.92\times 10^{-19}~J$
We can find the threshold wavelength for this work function:
$\phi = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{\phi}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{3.92\times 10^{-19}~J}$
$\lambda = 5.07\times 10^{-7}~m$
$\lambda = 507~nm$
The maximum wavelength of photons that will eject electrons from this metal is $507~nm$
