1.

The threshold wavelength of photo sensitive metal is 5000Å. Find the velocity of the photoelectrons emitted by it when radiation of wavelength 4000 Å is incident on it. Given h = 6.625xx10^(-34)Js, e = 1.6xx10.^(-19)C and mass of electron = 9.1xx10^(-34)kg.

Answer»

Solution :Given : `lambda_0 = 5000Å = 5000 xx 10^(-10)m`, V = ?
`lambda = 4000 Å= 4000 xx 10^(-19) m`
`h= 6.625 xx 10^(-34)` JS
`e=1.6 xx 10^(-19)C`
`m= 9.1 xx 10^(-31) kg`
From Einstein.s photoelectric equation `K_("max") = HV - hv_0`
` =hC[(1)/(lambda)-(lambda)/(lambda_0)]`
`= 6.625xx10|^(-34) xx3xx10^8[(1)/(4000xx10^(-10))-(1)/(5000xx10^(-10))]`
`=19.875xx10^9-26[(1)/(4xx10^(-7))-(1)/(5xx10^(-7))]`
`=(19.875xx10^(-26))/(10^(-7))[1/4-1/5]`
`=19.875xx10^(-19)[(5-4)/920)]`
`=(19.875xx10^(-19))/(20)`
`K_("max")=0.9937xx10^(-19)J`
We know that
`K_("max")=1/2mv_("max")^2` `v_("max")^2=(1.9874xx10^(-19))/(9.1xx10^(-31))`
`v^2=0.2183xx10^(12)`
`v=4.672xx10^5 MS^(-1)`


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