Light of frequency `8.41xx10^(14)Hz` is incident on a metal surface. Electrons with their maximum speed of `7.5xx10^(5)ms^(-1)` are ejected from the surface. Calculate the threshold frequency for photoemission of electrons. Also find the work function of the metal in electron volt `(eV)`. Given Plank's constant `h=6.625xx10^(-34)Js` and mass of the electron `9.1xx10^(-31)kg`.

Explanation

Given `f=8.41xx10^14` Hz.
`v=7.5xx10^5 ms^(-1)` .
`v_0`=? `omega`=?
`h=6.625xx10^(-34)` Js, `m=9.1xx10^(-31)` kg
Since `E=55.716xx10^(-20)` J
`=5.5716xx10^(-19)`J
Maximum `KE=1/2mv^2`
`=1/2xx9.1xx10^(-31)xx(7.5xx10^5)^2`
`=1/2xx9.1xx56.25xx10^(-21)`
`=255.937xx10^(-21)`J
`(KE)_(max)=2.559xx10^(-19)J`
Using P.E. equation , work function
`w=hv-(K.E)_(max)`
`therefore w=hv_0=(5.5716-2.559)xx10^(-19)` J
`therefore ` W=Work function `=hv_0=3.0126xx10^(-19)` J
Threshold frequency `v_0=(3.0126xx10^(-19))/(6.625xx10^(-34))`Hz
or `v_0=0.4547xx10^15` Hz
i.e. Threshold frequency `=4.547xx10^14` Hz
Work function =`(3.0126xx10^(-19))/(1.6xx10^(-19))`=1.883 eV
Work function = 1.883 eV