The threshold wavelength of a photosensitive metal is 662.5 nm. If this metal is irradiated with a radiation of wavelength 331.3 nm, find the maximum kinetic energy of the photoelectrons. If the wavelength of radiation is increased to 496.5 nm, calculate the change in maximum kinetic energy of the photoelectrons. (Planck's constant `h = 6.625 xx 10^(-34)` Js and `"speed of light in vacuum" = 3 xx 10^(8)ms^(-1)`)

Explanation

Given : `lambda_0=662.5xx10^(-9)` m
`lambda=331.3xx10^(-9)`m
`lambda.=496xx10^(-9) m `
`h=6.625xx10^(-3)4` Js
`C=3xx10^8` m/s
`K.E.=hc[1//lambda-1lambda_0]`
`=(6.625xx10^(-34))(3xx10^8)[1/(331.3xx10^(-9))-1/(662.5xx10^(-9))]`
`K.E.=(6.625xx10^(-34))(3xx10^8)[10^9/331.3-10^9/662.5]`
`K.E.=(6.625xx3)xx10^(-34+8+9)[1/331.3-1/662.5]`
`K.E.=19.87xx10^(-17)[0.0030-0.0015]`
`K.E.=19.87xx10^(-17)[0.0015]`
`K.E.=0.0298xx10^(-17)J =2.98xx10^(-19)J`
`K.E.=19.87xx10^(-17)[1/331-1/0.493]=19.87xx10^(-17)[0.0030-0.0020]`
`=19.87xx10^(-17)xx0.0010`
`K.E.=1.98xx10^(-19)` J .