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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] Find the ratio of the wavelenght of incident light to the De-Brogli wavelenght of the fastest photoelectron emitted

Answer»

`23.6`
`33.1`
`186.4`
`286.4`

Solution :WAVELENGHT of incident light `=lambda=(1240)/(5)nm`=`248nm` WAVELENGHTOF fastest photonsphotonelectrons`=lambda_(1)`
`lambda_(1)=(h)/sqrt(2m_(e) KE)=sqrt((150)/(2))`
`lambda_(1)=8.66=0.866 nm`
`lambda_(1)/lambda_(1)=(248)/(0.866)=286.374`


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