In a photoelectriceffect setup, a point source light of power ((p)/(4piR^(2))) =3.2xx10^(-3)W emitsmonoenergetic photons of enegry5.0 EV.Then photons falling on a metallic sphere of radius r=6 mm and of work function 3.0 eV at distance of 0.6 mm. The efficiency of photo- electron emissionis on for every 10^(5)incident photons.Assume that the sphere is located and initially neuttral and that photoelectron are instantly swept away after emission. [Given hc=1240 eV-nm] Calculate the number of photoelectron emitted second.

In a photoelectriceffect setup, a point source light of power ((p)/(4p

1.	In a photoelectriceffect setup, a point source light of power ((p)/(4piR^(2))) =3.2xx10^(-3)W emitsmonoenergetic photons of enegry5.0 EV.Then photons falling on a metallic sphere of radius r=6 mm and of work function 3.0 eV at distance of 0.6 mm. The efficiency of photo- electron emissionis on for every 10^(5)incident photons.Assume that the sphere is located and initially neuttral and that photoelectron are instantly swept away after emission. [Given hc=1240 eV-nm] Calculate the number of photoelectron emitted second.
Answer» `10^(10)` `2xx10^(8)` `5XX10^(6)` `10^(6)` Solution :Total number of photons/SECOND =n photos/second `nxx5xx1.6xx10^(-19)`=`3.2xx10^(-3)` `rArr=(3.2xx10^(-3))/(5xx1.6xx10^(-19))=4xx10^(15)` photons /second. number of photons falling per second on metallic sphere ='n' `n=nxx(pir)/(4piR^(2))""r=6mn=0.006mn,""R=0.6 m` `rArrn4xx10^(15)xx(0.006xx0.006)/(4xx0.6xx0.6)`photons /sec. `=10^(11)` photons /sec `therefore`number of photons electronnemitted /sec. `=(10^(11))/(10^(5))=10^(6)`