1.

Wavelength of light incident on a photo-sensitive surface is reduced from 3500 Å to 290 nm.The change in stopping potential is ……..(h=6.625xx10^(-34)Js)

Answer»

SOLUTION :`lambda=3500Å,lambda_(2)=290 nm.`
`h=6.625x10^(-34)` Js(given values)
`(1)/(2)mv_(max)^(2)=(HC)/(lambda)-phi=eV_(0)`
`therefore V_(01)e=(hc)/(lambda_(1))-phi`
`therefore V_(02)e=(hc)/(lambda_(2))-phi`
`therefore (V_(02)-V_(01))e=hc((1)/(lambda_(2))-(1)/(lambda_(1)))`
`therefore V_(02)-V_(01)=(hc)/(e)[(lambda_(1)-lambda_(2))/(lambda_(1)lambda_(2))]`
`=(6.625xx10^(-34)xx3xx10^(8))/(1.6xx10^(10^(-19)))`
`[(3.5xx10^(-7)-2.9xx10^(-7))/(3.5xx10^(-7)xx2.9xx10^(-7))]`
`=12.42[(0.6)/(3.5xx2.9)]`


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