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

$8 \times 10^{14} Hz$

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

Given,

ϕ, Work function = 3.3 eV

h, Planck's constant = $4.13×10^{-15}eV$

Now,

$f_0$, minimum frequency of incident light required to eject photoelectrons

$ϕ=h.f_0$

$f_0=\frac{ϕ}{h}=\frac{3.3}{4.1357×10^{-15}}$

$≈8 \times 10^{14} Hz$