Solve for wavelength of photon given threshold frequency The threshold frequency of a potassium metal plate is $\pu{4.35\times10^14 Hz}$. Determine the wavelength of a photon required to reach $\pu{2.50 eV}$ of the maximum kinetic energy of an emitted electron.  

Options: 
A. $\ \ \ \pu{2.88\times10^2 nm}$   
B. $\pu{-4.96\times10^2 nm}$   
C. $\ \ \ \pu{9.50\times10^-21 nm}$  
D. $\ \ \ \pu{6.69\times10^2 nm}$  
E. $\ \ \ \pu{2.02\times10^15 nm}$ 

From what I understand, I used the equation $KE = h\nu + h\nu_0$ and replaced $\nu$ with $\frac cA$ because wavelength and frequency are inversely proportional. Then I plugged in the given components, $\nu_0$ and a fraction of $KE$ to solve; however, I don't understand how to approach this as I'm only given $2.5 \pu{eV} (4.005 \times 10^{-19} \pu J)$ of maximum $KE$, not an actual value for $KE$. The answer I'm getting is $\pu{1.14 \times 10^{-12} nm}$ if I just use $\pu{2.5 eV} = KE$ max, which isn't an answer choice, and the given answer is A. $2.88 \times 10^2 \pu{nm}$. Please help.
Always work in energy not wavelength. You are given $\nu_0$ in Hz ($\mathrm{s}^{-1}$), multiply this by $h$ to convert it to Joules; ($h$ has units J s). You have the kinetic energy in eV, use the conversion  $1\mathrm{eV} = 1.602\cdot 10^{-19}$ J to get into joules. Now you have only to find energy $h\nu$ by subtraction. Finally convert energy to frequency using $E=h\nu$ and then to wavelength using $c=\lambda \nu$. Make sure $c$ is in m/s and as you answer will be in meters then convert to nm from metres.