The correct option is A 5×1014Hz
Using eVs=hcλ−ϕ
where ϕ is the work function of metal, hc=1241eV if wavelength is put in nm.
Case 1: λ=550nm,Vs=0.19V
∴0.19eV=1241550eV−ϕ
Or 0.19eV=2.256eV−ϕ
⇒ϕ=2.07eV
Threshold frequency for the surface
vth=ϕh=2.07×1.6×10−196.62×10−34=5×1014Hz