a b c d

b

$ K_{max} = \frac{hc}{\lambda} - \phi $ For First case $K_1= \frac{1240}{500} - \phi = 2.48eV - \phi$ For Second case $K_1= \frac{1240}{200} - \phi = 6.2eV - \phi$ $\Rightarrow 6.2eV - \phi = 3(2.48eV - \phi) = 7.44eV - 3\phi$ $\Rightarrow 2\phi = 1.24eV $ $\Rightarrow \phi = 0.62eV$