$ E_1 = \frac{hc}{\lambda_1} =\frac{1240}{350} = 3.54 eV$

$ E_2 = \frac{hc}{\lambda_2} =\frac{1240}{540} = 2.30 eV$

$V_{max} = \sqrt{\frac{2(E - \phi)}{m}}$

$\frac{v_1}{v_2} = \sqrt{\frac{3.54 - \phi}{2.30 - \phi}} = 4$

$ 3.54 - \phi = 9.20 - 4\phi$

$ 3\phi = 5.66 eV$

$ \phi = 1.9 eV$