The work function of a photoelectric surface is 3.3 eV. The value of the threshold frequency for the surface is:

$8 \times 10^{14} Hz$ $5 \times 10^{16} Hz$ $4 \times 10^{11} Hz$ $8 \times 10^{10} Hz$

$8 \times 10^{14} Hz$

The correct answer is Option (1) → $8 \times 10^{14} Hz$