The electric field associated with a light wave is $$E=E0sin$$⁡$$1$$⋅$$57$$×$$107(x$$−$$ct)$$ where x is in metre and t is in second. If this light is used to produce $$photo-electric$$ emission from the surface of a metal of $$work-function$$ $$1$$'$$9$$ eV, then the stopping potential will be

Question

The electric field associated with a light wave is $$E=E0sin$$⁡$$1$$⋅$$57$$×$$107(x$$−$$ct)$$ where x is in metre and t is in second. If this light is used to produce $$photo-electric$$ emission from the surface of a metal of $$work-function$$ $$1$$'$$9$$ eV, then the stopping potential will be

Infinity Learn · Accepted Answer

Given $$thatE=E0sin$$⁡$$1$$⋅$$57$$×$$107(x$$−$$ct)$$  We know that, $$E=E0sin$$⁡$$2$$πλ(x$$−$$ct)  Comparing, we get λ$$=2$$$$π$$$$1$$⋅$$57$$×$$107m$$ ∴  Energy of photon $$=hc$$λ $$=6$$⋅$$64$$×$$10$$−$$343$$×$$1081$$⋅$$57$$×$$1072$$π×$$1$$⋅$$6$$×$$10$$−$$19eV$$ $$=3$$⋅$$1eV$$ So, maximum K.E. of $$photo-electron$$ $$=3$$⋅$$1$$−$$1$$⋅$$9=1$$⋅$$2eV$$ This is equal to the stopping potential.

The electric field associated with a light wave is $E = E_{0} \sin [1 \cdot 57 \times 10^{7} (x - c t)]$ where x is in metre and t is in second. If this light is used to produce photo-electric emission from the surface of a metal of work-function 1'9 eV, then the stopping potential will be

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The electric field associated with a light wave is E=E0sin⁡1⋅57×107(x−ct) where x is in metre and t is in second. If this light is used to produce photo-electric emission from the surface of a metal of work-function 1'9 eV, then the stopping potential will be

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The electric field associated with a light wave is $E = E_{0} \sin [1 \cdot 57 \times 10^{7} (x - c t)]$ where x is in metre and t is in second. If this light is used to produce photo-electric emission from the surface of a metal of work-function 1'9 eV, then the stopping potential will be