The work function of a photoelectric surface is 3.3 eV. The value of the threshold frequency for the surface is:
Answer & explanation
Correct answer: option 1
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
$ϕ=h.f_0$
$f_0=\frac{ϕ}{h}=\frac{3.3}{4.1357×10^{-15}}$
$≈8 \times 10^{14} Hz$