The variation of stopping potential (V0) as a function of the frequency (v) of the incident light for a metal is shown in figure. The work function of the surface is
(A)18.6 eV
(B)2.98 eV
(C)2.07 eV✓ Correct
(D)1.36 eV
Answer: (C)
Solution
From the photoelectric equation, eV0=hν−ϕ. At the threshold frequency (where V0=0), we have ϕ=hν0. From the graph, the threshold frequency is ν0=5×1014 Hz. Therefore, ϕ=6.6×10−34×5×1014=33×10−20 J =1.6×10−1933×10−20=2.07 eV.
JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Single correct option