E=10−3cos[5×10−72πx−2π×6×1014t]x^N/c
k=5×10−72π
And ω=6×1014×2π ⇒f=6×1014Hz
E=ϕ+KEmax
KEmax=hf−ϕ
=1.6×10−196.6×10−34×6×1014−2
=2.475−2=0.48eV
∴ Stopping potential is 0.48V
JEE Main 2019 — Physics Modern Physics
The electric field of light wave is given as \vec{E }={10}^{–3}\mathrm{cos}(\frac{2\pi x}{5\times {10}^{–7}}-2\pi \times 6\times {10}^{14} t)\overset{^}{x}\frac{N}{C} . This light falls on a metal plate of work function 2eV . The stopping potential of the photo-electrons is:
Given, E (in eV ) =λ(inA˚)12375
Held on 9 Apr 2019 · Verified 6 Jul 2026.
0.72V
2.0V
2.48V
0.48V
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