Answer
$(B)~1.08eV$
Work Step by Step
We know that for a lithium metal surface $W_{\circ}=2.29eV(1.6\times 10^{-19}J/eV)$
$W_{\circ}=3.664\times 10^{-19}J$
Now $K.E_{max}=\frac{hc}{\lambda}-W_{\circ}$
We plug in the known values to obtain:
$K.E_{max}=\frac{(6.57\times 10^{-34}J.s)(3.0\times 10^8m/s)}{365.0\times 10^{-9}}-3.664\times 10^{-19}J$
$K.E_{max}=1.736\times 10^{-19}J$
$K.E_{max}=\frac{1.736\times 10^{-19}J}{1.6\times 10^{-19}J/eV}$
$K.E_{max}=1.08eV$
