Two metallic plates A and B, each of area $$5$$ x $$10-4$$ $$m2$$ are placed parallel to each other at a separation of $$1$$ cm. Plate B carries a positive charge of $$33.7$$ pC. A monochromatic beam of light, with photons of energy $$5$$ eV each, starts falling on plate A at t $$=$$ $$0$$, so that $$1016$$ photons fall on it per square meter per second. Assume that one photoelectron is emitted for every $$106$$ incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value $$2$$ eV. Electric field between the plates at the end of $$10$$ seconds is

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Infinity Learn · Accepted Answer

Number of photoelectrons emitted up to t $$=$$ $$10$$ $$sec$$ $$aren=$$ (Number$$ of photons per unit area per unit time×$$($$ Area $$)×$$($$ Time $$)$$ $$106=1106(10)16$$×$$5$$×$$10$$−$$4$$×(10)=5$$×$$107At$$ time t $$=10sPhotoelectrons$$ are ejected from plate A, so it will acquire positive charge. Charge on plate A : $$qA=+ne=5$$×$$107$$×$$1.6$$×$$10$$−$$19=8$$×$$10$$−$$12C=8pCPhotoelectrons$$ are absorbed by plate B, so its initial charge reduces.Charge on plate B : $$qB=33.7$$−$$8=25.7pCConsider$$ these plates as of capacitor.Charge on sides facing each other $$=qB-qA2Electric$$ field between the plates $$E=$$σε$$0=qB$$−$$qA2$$ε$$0A=(25.7$$−$$8)×$$10$$−$$122$$×$$8.85$$×$$10$$−$$12$$×$$5$$×$$10$$−$$4=2$$×$$103NC$$

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