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The electric field associated with a light wave is given by E=sin1.57×107m−1x−ct. If this light is used in an experiment on photoelectric effect .The stopping potential is n×0.2 volt. Then n is, ( The emitter having work function 1.9 eV and consider hc=12400 eV−A0)

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Detailed Solution

By comparing given equation k=2πλ⇒1.5×107=2×3.14λ⇒λ=4×10−7m⇒λ=4000A0Now, KE=E−W0⇒eV0=12400λ−1.9=(3.1−1.9)eV⇒V0=1.2 volt

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