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The work function of a photosensitive element is 2eV. Calculate the velocity of a photoelectron when the element is exposed to a light of wavelength 4xx10^(3) A.

Answer»

Solution :Einstein.s photoelectric equation is `(1)/(2)mv^(2)=(HC)/(lamda)-W_(0)`
`(1)/(2)mv^(2)=(6.62xx3)/(4xx10^(3)xx10^(-10))xx10^(-26)-2xx1.6xx10^(-19)`
`v^(2)=(1.765xx2)/(9.1)xx10^(12)`
`v=sqrt((1.765xx2)/(9.1))xx10^(6)=6.228xx10^(5)ms^(-1)`


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