If the work function of the photoelectric surface is 3.3 eV, then the value of threshold frequency is:
Answer & explanation
Correct answer: option 1
Here, $\phi_0=3.3eV=3.3×1.6×10^{-19}J$
Work function, $\phi_0=hv_0$
Where $v_0$ is the threshold frequency $v_0=\frac{\phi_0}{h}=\frac{3.3×1.6×10^{-19}J}{6.6×10^{-34}Js}=0.8×10^{15}Hz=8×10^{14}Hz$