Answer
The stopping potential is $~~2.00~V$
Work Step by Step
We can find the photon energy:
$E = \frac{hc}{\lambda}$
$E = \frac{(6.63\times 10^{-34}~J\cdot s)(3.0\times 10^8~m/s)}{200\times 10^{-9}~m}$
$E = 9.945\times 10^{-19}~J$
$E = (9.945\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 6.20~eV$
We can find the maximum kinetic energy of the ejected electrons:
$E = K_{max}+ \Phi$
$K_{max} = E -\Phi$
$K_{max} = (6.20~eV) -(4.20~eV)$
$K_{max} = 2.00~eV$
We can find the stopping potential:
$(1~e)(\Delta V) = 2.00~eV$
$\Delta V = 2.00~V$
The stopping potential is $~~2.00~V$
