Answer
The stopping potential is $~~1.3~V$
Work Step by Step
We can find the photon energy:
$E = \frac{hc}{\lambda}$
$E = \frac{(6.63\times 10^{-34}~J\cdot s)(3.0\times 10^8~m/s)}{400\times 10^{-9}~m}$
$E = 4.9725\times 10^{-19}~J$
$E = (4.9725\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 3.1~eV$
We can find the stopping potential:
$E = K_{max} +\Phi$
$K_{max} = E-\Phi$
$(1~e)~\Delta V = E-\Phi$
$(1~e)~\Delta V = 3.1~eV-1.8~eV$
$(1~e)~\Delta V = 1.3~eV$
$\Delta V = 1.3~V$
The stopping potential is $~~1.3~V$
