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Question

# The electric field associated with a light wave is given by E=E0sin[(1.57×107)(x−ct)](where, x and t are in metre and second). The stopping potential when its light is used in an experiment on photoelectric effect with the emitter having work function ϕ=19eV is?

A
0.6eV
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B
1.2eV
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C
1.8eV
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D
2.4eV
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Solution

## The correct option is D 1.2eVGiven, E=E0sin(1.57×107)(x−ct) ........(i)∴E=E0sinωC(x−ct) .........(ii)On comparing Eqs. (i) and (ii), we getωC=1.57×107=π2×107λ=4×10−7m=400nmAs, ϕ0=hcλ−Ek1.9=1242400−EkEk=(3.1−1.9)eV=1.2eV.

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