When a beam of $$10.6$$ eV photon of intensity $$2$$ $$w/m2$$ falls on a platinum surface of area $$104$$ $$m2$$ and work function $$5.6$$ eV, $$0.53$$% of incident photons eject photoelectrons. The number of photoelectrons emitted per second is $$125$$ n×$$1017$$. Calculate n

Question

When a beam of $$10.6$$ eV photon of intensity $$2$$  $$w/m2$$ falls on a platinum surface of area  $$104$$ $$m2$$ and work function  $$5.6$$  eV, $$0.53$$% of incident photons eject photoelectrons. The number of photoelectrons emitted per second is  $$125$$ n×$$1017$$. Calculate n

Infinity Learn · Accepted Answer

Power  $$=2$$×$$104$$  WattsNo of photons falling per second $$=$$ $$MM=2$$×$$10410.6$$×$$1.6$$×$$10$$−$$19$$ No of electrons ejected $$=N=0.53100$$×$$M=0.53100$$×$$2$$×$$10410.6$$×$$1.6$$×$$10$$−$$19N=625$$×$$1017=n$$×$$125$$×$$1017n=5$$

Photoelectric effect and study of photoelectric effect

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When a beam of 10.6 eV photon of intensity $2 w / m^{2}$ falls on a platinum surface of area $10^{4} m^{2}$ and work function $5.6 e V$ , 0.53% of incident photons eject photoelectrons. The number of photoelectrons emitted per second is $125 n \times 10^{17}$ . Calculate n

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Photoelectric effect and study of photoelectric effect

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When a beam of 10.6 eV photon of intensity 2 w/m2 falls on a platinum surface of area 104 m2 and work function 5.6 eV, 0.53% of incident photons eject photoelectrons. The number of photoelectrons emitted per second is 125 n×1017. Calculate n

Power =2×104 WattsNo of photons falling per second = MM=2×10410.6×1.6×10−19 No of electrons ejected =N=0.53100×M=0.53100×2×10410.6×1.6×10−19N=625×1017=n×125×1017n=5

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When a beam of 10.6 eV photon of intensity $2 w / m^{2}$ falls on a platinum surface of area $10^{4} m^{2}$ and work function $5.6 e V$ , 0.53% of incident photons eject photoelectrons. The number of photoelectrons emitted per second is $125 n \times 10^{17}$ . Calculate n