The electric field of light wave is given as \E^{\to} =10^{-3} \cos ( {2 \pi x}/{5 \times 10^{-7}}-2 \\pi\times 6 \\times 10^{14} t ) x^{\^} \frac{N}{C} . The light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is:Given, E (in eV) = {12375}/{ \lambda(i n \A ^{o} )} [9 Apr 2019 Shift 1]

JEE Main 9 Apr 2019 Paper 1 Solved Paper

Section: Physics

Question : 10 of 90

Marks: +1, -0

The electric field of light wave is given as

→

=10−3‌cos‌ (

2πx

5 ×10−7

−2π×6×1014t )‌

‌

. The light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is:Given, E (in eV) =

12375

λ(in

)

[9 Apr 2019 Shift 1]

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