Solution:
Solution
Energy of incident light
$$E\left( eV \right) =\dfrac { 12375 }{ 3320 } =3.72eV$$
We know that, $$E={ W }_{ 0 }+{ eV }_{ 0 }$$
$${ V }_{ 0 }=\dfrac { \left( E-{ W }_{ 0 } \right) }{ e } =\dfrac { 3.72eV-1.07eV }{ e } $$
$${ V }_{ 0 }=2.65$$ Volt